[英] 50 个 Go 开发者常犯的错误 - 推酷
[英] 50 个 Go 开发者常犯的错误
Go is a simple and fun language, but, like any other language, it has a few gotchas... Many of those gotchas are not entirely Go's fault. Some of these mistakes are natural traps if you are coming from another language. Others are due to faulty assumptions and missing details.
A lot of these gotchas may seem obvious if you took the time to learn the language reading the official spec, wiki, mailing list discussions, many great posts and presentations by Rob Pike, and the source code. Not everybody starts the same way though and that's OK. If you are new to Go the information here will save you hours debugging your code.
Total Beginner:
Advanced Beginner:
Opening Brace Can't Be Placed on a Separate Line
level: beginner
In most other languages that use braces you get to choose where you place them. Go is different. You can thank automatic semicolon injection (without lookahead) for this behavior. Yes, Go does have semicolons :-)
package main
import &fmt&
func main()
{ //error, can't have the opening brace on a separate line
fmt.Println(&hello there!&)
Compile Error:
/tmp/sandbox/main.go:6: syntax error: unexpected semicolon or newline before {
package main
import &fmt&
func main() {
fmt.Println(&works!&)
Unused Variables
level: beginner
If you have an unused variable your code will fail to compile. There's an exception though. You must use variables you declare inside functions, but it's OK if you have unused global variables.
If you assign a new value to the unused variable your code will still fail to compile. You need to use the variable value somehow to make the compiler happy.
package main
var gvar int //not an error
func main() {
var one int
//error, unused variable
//error, unused variable
var three int //error, even though it's assigned 3 on the next line
Compile Errors:
/tmp/sandbox/main.go:6: one declared and not used /tmp/sandbox/main.go:7: two declared and not used /tmp/sandbox/main.go:8: three declared and not used
package main
import &fmt&
func main() {
var one int
fmt.Println(two)
var three int
one = three
var four int
four = four
Another option is to comment out or remove the unused variables :-)
Unused Imports
level: beginner
Your code will fail to compile if you import a package without using any of its exported functions, interfaces, structures, or variables.
If you really need the imported package you can use the blank identifier,
, as its package name to avoid this compilation failure. The blank identifier is used to import packages for their side effects.
package main
func main() {
Compile Errors:
/tmp/sandbox/main.go:4: imported and not used: &fmt& /tmp/sandbox/main.go:5: imported and not used: &log& /tmp/sandbox/main.go:6: imported and not used: &time&
package main
var _ = log.Println
func main() {
_ = time.Now
Another option is to remove or comment out the unused imports :-)
Short Variable Declarations Can Be Used Only Inside Functions
level: beginner
package main
myvar := 1 //error
func main() {
Compile Error:
/tmp/sandbox/main.go:3: non-declaration statement outside function body
package main
var myvar = 1
func main() {
Redeclaring Variables Using Short Variable Declarations
level: beginner
You can't redeclare a variable in a standalone statement, but it is allowed in multi-variable declarations where at least one new variable is also declared.
The redeclared variable has to be in the same block or you'll end up with a shadowed variable.
package main
func main() {
one := 1 //error
Compile Error:
/tmp/sandbox/main.go:5: no new variables on left side of :=
package main
func main() {
one, two := 1,2
one,two = two,one
Accidental Variable Shadowing
level: beginner
The short variable declaration syntax is so convenient (especially for those coming from a dynamic language) that it's easy to treat it like a regular assignment operation. If you make this mistake in a new code block there will be no compiler error, but your app will not do what you expect.
package main
import &fmt&
func main() {
fmt.Println(x)
//prints 1
fmt.Println(x) //prints 1
fmt.Println(x) //prints 2
fmt.Println(x)
//prints 1 (bad if you need 2)
This is a very common trap even for experienced Go developers. It's easy to make and it could be hard to spot.
Can't Use &nil& to Initialize a Variable Without an Explicit Type
level: beginner
The &nil& identifier can be used as the &zero value& for interfaces, functions, pointers, maps, slices, and channels. If you don't specify the variable type the compiler will fail to compile your code because it can't guess the type.
package main
func main() {
var x = nil //error
Compile Error:
/tmp/sandbox/main.go:4: use of untyped nil
package main
func main() {
var x interface{} = nil
Using &nil& Slices and Maps
level: beginner
It's OK to add items to a &nil& slice, but doing the same with a map will produce a runtime panic.
package main
func main() {
var s []int
s = append(s,1)
package main
func main() {
var m map[string]int
m[&one&] = 1 //error
Map Capacity
level: beginner
You can specify the map capacity when it's created, but you can't use the
function on maps.
package main
func main() {
m := make(map[string]int,99)
cap(m) //error
Compile Error:
/tmp/sandbox/main.go:5: invalid argument m (type map[string]int) for cap
Strings Can't Be &nil&
level: beginner
This is a gotcha for developers who are used to assigning &nil& identifiers to string variables.
package main
func main() {
var x string = nil //error
if x == nil { //error
x = &default&
Compile Errors:
/tmp/sandbox/main.go:4: cannot use nil as type string in assignment /tmp/sandbox/main.go:6: invalid operation: x == nil (mismatched types string and nil)
package main
func main() {
var x string //defaults to && (zero value)
if x == && {
x = &default&
Array Function Arguments
level: beginner
If you are a C or C++ developer arrays for you are pointers. When you pass arrays to functions the functions reference the same memory location, so they can update the original data. Arrays in Go are values, so when you pass arrays to functions the functions get a copy of the original array data. This can be a problem if you are trying to update the array data.
package main
import &fmt&
func main() {
x := [3]int{1,2,3}
func(arr [3]int) {
arr[0] = 7
fmt.Println(arr) //prints [7 2 3]
fmt.Println(x) //prints [1 2 3] (not ok if you need [7 2 3])
If you need to update the original array data use array pointer types.
package main
import &fmt&
func main() {
x := [3]int{1,2,3}
func(arr *[3]int) {
(*arr)[0] = 7
fmt.Println(arr) //prints &[7 2 3]
fmt.Println(x) //prints [7 2 3]
Another option is to use slices. Even though your function gets a copy of the slice variable it still references the original data.
package main
import &fmt&
func main() {
x := []int{1,2,3}
func(arr []int) {
arr[0] = 7
fmt.Println(arr) //prints [7 2 3]
fmt.Println(x) //prints [7 2 3]
Unexpected Values in Slice and Array &range& Clauses
level: beginner
This can happen if you are used to the &for-in& or &foreach& statements in other languages. The &range& clause in Go is different. It generates two values: the first value is the item index while the second value is the item data.
package main
import &fmt&
func main() {
x := []string{&a&,&b&,&c&}
for v := range x {
fmt.Println(v) //prints 0, 1, 2
package main
import &fmt&
func main() {
x := []string{&a&,&b&,&c&}
for _, v := range x {
fmt.Println(v) //prints a, b, c
Slices and Arrays Are One-Dimensional
level: beginner
It may seem like Go supports multi-dimensional arrays and slices, but it doesn't. Creating arrays of arrays or slices of slices is possible though. For numerical computation apps that rely on dynamic multi-dimensional arrays it's far from ideal in terms of performance and complexity.
You can build dynamic multi-dimensional arrays using raw one-dimensional arrays, slices of &independent& slices, and slices of &shared data& slices.
If you are using raw one-dimensional arrays you are responsible for indexing, bounds checking, and memory reallocations when the arrays need to grow.
Creating a dynamic multi-dimensional array using slices of &independent& slices is a two step process. First, you have to create the outer slice. Then, you have to allocate each inner slice. The inner slices are independent of each other. You can grow and shrink them without affecting other inner slices.
package main
func main() {
table := make([][]int,x)
for i:= range table {
table[i] = make([]int,y)
Creating a dynamic multi-dimensional array using slices of &shared data& slices is a three step process. First, you have to create the data &container& slice that will hold raw data. Then, you create the outer slice. Finally, you initialize each inner slice by reslicing the raw data slice.
package main
import &fmt&
func main() {
h, w := 2, 4
raw := make([]int,h*w)
for i := range raw {
raw[i] = i
fmt.Println(raw,&raw[4])
//prints: [0 1 2 3 4 5 6 7] &ptr_addr_x&
table := make([][]int,h)
for i:= range table {
table[i] = raw[i*w:i*w + w]
fmt.Println(table,&table[1][0])
//prints: [[0 1 2 3] [4 5 6 7]] &ptr_addr_x&
There's a spec/proposal for multi-dimensional arrays and slices, but it looks like it's a low priority feature at this point in time.
Accessing Non-Existing Map Keys
level: beginner
This is a gotcha for developers who expect to get &nil& identifiers (like it's done in other languages). The returned value will be &nil& if the &zero value& for the corresponding data type is &nil&, but it'll be different for other data types. Checking for the appropriate &zero value& can be used to determine if the map record exists, but it's not always reliable (e.g., what do you do if you have a map of booleans where the &zero value& is false). The most reliable way to know if a given map record exists is to check the second value returned by the map access operation.
package main
import &fmt&
func main() {
x := map[string]string{&one&:&a&,&two&:&&,&three&:&c&}
if v := x[&two&]; v == && { //incorrect
fmt.Println(&no entry&)
package main
import &fmt&
func main() {
x := map[string]string{&one&:&a&,&two&:&&,&three&:&c&}
if _,ok := x[&two&]; !ok {
fmt.Println(&no entry&)
Strings Are Immutable
level: beginner
Trying to update an individual character in a string variable using the index operator will result in a failure. Strings are read-only byte slices (with a few extra properties). If you do need to update a string then use a byte slice instead converting it to a string type when necessary.
package main
import &fmt&
func main() {
x := &text&
x[0] = 'T'
fmt.Println(x)
Compile Error:
/tmp/sandbox/main.go:7: cannot assign to x[0]
package main
import &fmt&
func main() {
x := &text&
xbytes := []byte(x)
xbytes[0] = 'T'
fmt.Println(string(xbytes)) //prints Text
Note that this isn't really the right way to update characters in a text string because a given character could be stored in multiple bytes. If you do need to make updates to a text string convert it to a rune sclice first. Even with rune slices a single character might span multiple runes, which can happen if you have characters with grave accent, for example. This complicated and ambiguous nature of &characters& is the reason why Go strings are represented as byte sequences.
Conversions Between Strings and Byte Slices
level: beginner
When you convert a string to a byte slice (and vice versa) you get a complete copy of the orginal data. It's not like a cast operation in other languages and it's not like reslicing where the new slice variable points to the same underlying array used by the original byte slice.
Go does have a couple of optimizations for
conversions to avoid extra allocations (with more optimizations on the todo list).
The first optimization avoids extra allocations when
keys are used to lookup entries in
map[string]
collections:
m[string(key)]
The second optimization avoids extra allocations in
clauses where strings are converted to
for i,v := range []byte(str) {...}
Strings and Index Operator
level: beginner
The index operator on a string returns a byte value, not a character (like it's done in other languages).
package main
import &fmt&
func main() {
x := &text&
fmt.Println(x[0]) //print 116
fmt.Printf(&%T&,x[0]) //prints uint8
If you need to access specific string &characters& (unicode code points/runes) use the
clause. The official &unicode/utf8& package and the experimental utf8string package (golang.org/x/exp/utf8string) are also useful. The utf8string package includes a convenient
method. Converting the string to a slice of runes is an option too.
Strings Are Not Always UTF8 Text
level: beginner
String values are not required to be UTF8 text. They can contain arbitrary bytes. The only time strings are UTF8 is when string literals are used. Even then they can include other data using escape sequences.
To know if you have a UTF8 text string use the
ValidString()
function from the &unicode/utf8& package.
package main
&unicode/utf8&
func main() {
data1 := &ABC&
fmt.Println(utf8.ValidString(data1)) //prints: true
data2 := &A\xfeC&
fmt.Println(utf8.ValidString(data2)) //prints: false
String Length
level: beginner
Let's say you are a python developer and you have the following piece of code:
data = u'?'
print(len(data)) #prints: 1
When you convert it to a similar Go code snippet you might be surprised.
package main
import &fmt&
func main() {
data := &?&
fmt.Println(len(data)) //prints: 3
The built-in
function returns the number of bytes instead of the number of characters like it's done for unicode strings in Python.
To get the same results in Go use the
RuneCountInString()
function from the &unicode/utf8& package.
package main
&unicode/utf8&
func main() {
data := &?&
fmt.Println(utf8.RuneCountInString(data)) //prints: 1
Technically the
RuneCountInString()
function doesn't return the number of characters because a single character may span multiple runes.
package main
&unicode/utf8&
func main() {
data := &e?&
fmt.Println(len(data))
//prints: 3
fmt.Println(utf8.RuneCountInString(data)) //prints: 2
Missing Comma In Multi-Line Slice, Array, and Map Literals
level: beginner
package main
func main() {
x := []int{
Compile Errors:
/tmp/sandbox/main.go:6: syntax error: need trailing comma before newline in composite literal /tmp/sandbox/main.go:8: non-declaration statement outside function body /tmp/sandbox/main.go:9: syntax error: unexpected }
package main
func main() {
x := []int{
y := []int{3,4,} //no error
You won't get a compiler error if you leave the trailing comma when you collapse the declaration to be on a single line.
log.Fatal and log.Panic Do More Than Log
level: beginner
Logging libraries often provide different log levels. Unlike those logging libraries, the log package in Go does more than log if you call its
functions. When your app calls those functions Go will also terminate your app :-)
package main
import &log&
func main() {
log.Fatalln(&Fatal Level: log entry&) //app exits here
log.Println(&Normal Level: log entry&)
Built-in Data Structure Operations Are Not Synchronized
level: beginner
Even though Go has a number of features to support concurrency natively, concurrency safe data collections are not one them :-) It's your responsibility to ensure the data collection updates are atomic. Goroutines and channels are the recommended way to implement those atomic operations, but you can also leverage the &sync& package if it makes sense for your application.
Iteration Values For Strings in &range& Clauses
level: beginner
The index value (the first value returned by the &range& operation) is the index of the first byte for the current &character& (unicode code point/rune) returned in the second value. It's not the index for the current &character& like it's done in other languages. Note that an actual character might be represented by multiple runes. Make sure to check out the &norm& package (golang.org/x/text/unicode/norm) if you need to work with characters.
clauses with string variables will try to interpret the data as UTF8 text. For any byte sequences it doesn't understand it will return 0xfffd runes (aka unicode replacement characters) instead of the actual data. If you have arbitrary (non-UTF8 text) data stored in your string variables, make sure to convert them to byte slices to get all stored data as is.
package main
import &fmt&
func main() {
data := &A\xfe\x02\xff\x04&
for _,v := range data {
fmt.Printf(&%#x &,v)
//prints: 0x41 0xfffd 0x2 0xfffd 0x4 (not ok)
fmt.Println()
for _,v := range []byte(data) {
fmt.Printf(&%#x &,v)
//prints: 0x41 0xfe 0x2 0xff 0x4 (good)
Iterating Through a Map Using a &for range& Clause
level: beginner
This is a gotcha if you expect the items to be in a certain order (e.g., ordered by the key value). Each map iteration will produce different results. The Go runtime tries to go an extra mile randomizing the iteration order, but it doesn't always succeed so you may get several identical map iterations. Don't be surprised to see 5 identical iterations in a row.
package main
import &fmt&
func main() {
m := map[string]int{&one&:1,&two&:2,&three&:3,&four&:4}
for k,v := range m {
fmt.Println(k,v)
And if you use the Go Playground (
) you'll always get the same results because it doesn't recompile the code unless you make a change.
Fallthrough Behavior in &switch& Statements
level: beginner
The &case& blocks in &switch& statements break by default. This is different from other languages where the default behavior is to fall through to the next &case& block.
package main
import &fmt&
func main() {
isSpace := func(ch byte) bool {
switch(ch) {
case ' ': //error
case '\t':
return true
return false
fmt.Println(isSpace('\t')) //prints true (ok)
fmt.Println(isSpace(' '))
//prints false (not ok)
You can force the &case& blocks to fall through by using the &fallthrough& statement at the end of each &case& block. You can also rewrite your switch statement to use expression lists in the &case& blocks.
package main
import &fmt&
func main() {
isSpace := func(ch byte) bool {
switch(ch) {
case ' ', '\t':
return true
return false
fmt.Println(isSpace('\t')) //prints true (ok)
fmt.Println(isSpace(' '))
//prints true (ok)
Increments and Decrements
level: beginner
Many languages have increment and decrement operators. Unlike other languages, Go doesn't support the prefix version of the operations. You also can't use these two operators in expressions.
package main
import &fmt&
func main() {
data := []int{1,2,3}
++i //error
fmt.Println(data[i++]) //error
Compile Errors:
/tmp/sandbox/main.go:8: syntax error: unexpected ++ /tmp/sandbox/main.go:9: syntax error: unexpected ++, expecting :
package main
import &fmt&
func main() {
data := []int{1,2,3}
fmt.Println(data[i])
Bitwise NOT Operator
level: beginner
Many languages use
as the unary NOT operator (aka bitwise complement), but Go reuses the XOR operator (
) for that.
package main
import &fmt&
func main() {
fmt.Println(~2) //error
Compile Error:
/tmp/sandbox/main.go:6: the bitwise complement operator is ^
package main
import &fmt&
func main() {
var d uint8 = 2
fmt.Printf(&%08b\n&,^d)
Go still uses
as the XOR operator, which may be confusing for some people.
If you want you can represent a unary NOT operation (e.g,
) with a binary XOR operation (e.g.,
0x02 XOR 0xff
). This could explain why
is reused to represent unary NOT operations.
Go also has a special 'AND NOT' bitwise operator (
), which adds to the NOT operator confusion. It looks like a special feature/hack to support
A AND (NOT B)
without requiring parentheses.
package main
import &fmt&
func main() {
var a uint8 = 0x82
var b uint8 = 0x02
fmt.Printf(&%08b [A]\n&,a)
fmt.Printf(&%08b [B]\n&,b)
fmt.Printf(&%08b (NOT B)\n&,^b)
fmt.Printf(&%08b ^ %08b = %08b [B XOR 0xff]\n&,b,0xff,b ^ 0xff)
fmt.Printf(&%08b ^ %08b = %08b [A XOR B]\n&,a,b,a ^ b)
fmt.Printf(&%08b & %08b = %08b [A AND B]\n&,a,b,a & b)
fmt.Printf(&%08b &^%08b = %08b [A 'AND NOT' B]\n&,a,b,a &^ b)
fmt.Printf(&%08b&(^%08b)= %08b [A AND (NOT B)]\n&,a,b,a & (^b))
Operator Precedence Differences
level: beginner
Aside from the &bit clear& operators (
) Go has a set of standard operators shared by many other languages. The operator precedence is not always the same though.
package main
import &fmt&
func main() {
fmt.Printf(&0x2 & 0x2 + 0x4 -& %#x\n&,0x2 & 0x2 + 0x4)
//prints: 0x2 & 0x2 + 0x4 -& 0x6
//Go: (0x2 & 0x2) + 0x4
//C++: 0x2 & (0x2 + 0x4) -& 0x2
fmt.Printf(&0x2 + 0x2 && 0x1 -& %#x\n&,0x2 + 0x2 && 0x1)
//prints: 0x2 + 0x2 && 0x1 -& 0x6
0x2 + (0x2 && 0x1)
(0x2 + 0x2) && 0x1 -& 0x8
fmt.Printf(&0xf | 0x2 ^ 0x2 -& %#x\n&,0xf | 0x2 ^ 0x2)
//prints: 0xf | 0x2 ^ 0x2 -& 0xd
//Go: (0xf | 0x2) ^ 0x2
//C++: 0xf | (0x2 ^ 0x2) -& 0xf
Unexported Structure Fields Are Not Encoded
level: beginner
The struct fields starting with lowercase letters will not be (json, xml, gob, etc.) encoded, so when you decode the structure you'll end up with zero values in those unexported fields.
package main
&encoding/json&
type MyData struct {
two string
func main() {
in := MyData{1,&two&}
fmt.Printf(&%#v\n&,in) //prints main.MyData{One:1, two:&two&}
encoded,_ := json.Marshal(in)
fmt.Println(string(encoded)) //prints {&One&:1}
var out MyData
json.Unmarshal(encoded,&out)
fmt.Printf(&%#v\n&,out) //prints main.MyData{One:1, two:&&}
App Exits With Active Goroutines
level: beginner
The app will not wait for all your goroutines to complete. This is a common mistake for beginners in general. Everybody starts somewhere, so there's no shame in making rookie mistakes :-)
package main
func main() {
workerCount := 2
for i := 0; i & workerC i++ {
go doit(i)
time.Sleep(1 * time.Second)
fmt.Println(&all done!&)
func doit(workerId int) {
fmt.Printf(&[%v] is running\n&,workerId)
time.Sleep(3 * time.Second)
fmt.Printf(&[%v] is done\n&,workerId)
You'll see:
[0] is running
[1] is running
One of the most common solutions is to use a &WaitGroup& variable. It will allow the main goroutine to wait until all worker goroutines are done. If your app has long running workers with message processing loops you'll also need a way to signal those goroutines that it's time to exit. You can send a &kill& message to each worker. Another option is to close a channel all workers are receiving from. It's a simple way to signal all goroutines at once.
package main
func main() {
var wg sync.WaitGroup
done := make(chan struct{})
workerCount := 2
for i := 0; i & workerC i++ {
go doit(i,done,wg)
close(done)
fmt.Println(&all done!&)
func doit(workerId int,done &-chan struct{},wg sync.WaitGroup) {
fmt.Printf(&[%v] is running\n&,workerId)
defer wg.Done()
fmt.Printf(&[%v] is done\n&,workerId)
If you run this app you'll see:
[0] is running
[0] is done
[1] is running
[1] is done
Looks like the workers are done before the main goroutine exists. Great! However, you'll also see this:
fatal error: all goroutines are asleep - deadlock!
That's not so great :-) What's going on? Why is there a deadlock? The workers exited and they executed
. The app should work.
The deadlock happens because each worker gets a copy of the original &WaitGroup& variable. When workers execute
it has no effect on the &WaitGroup& variable in the main goroutine.
package main
func main() {
var wg sync.WaitGroup
done := make(chan struct{})
wq := make(chan interface{})
workerCount := 2
for i := 0; i & workerC i++ {
go doit(i,wq,done,&wg)
for i := 0; i & workerC i++ {
close(done)
fmt.Println(&all done!&)
func doit(workerId int, wq &-chan interface{},done &-chan struct{},wg *sync.WaitGroup) {
fmt.Printf(&[%v] is running\n&,workerId)
defer wg.Done()
case m := &- wq:
fmt.Printf(&[%v] m =& %v\n&,workerId,m)
case &- done:
fmt.Printf(&[%v] is done\n&,workerId)
Now it works as expected :-)
Sending to an Unbuffered Channel Returns As Soon As the Target Receiver Is Ready
level: beginner
The sender will not be blocked until your message is processed by the receiver. Depending on the machine where you are running the code, the receiver goroutine may or may not have enough time to process the message before the sender continues its execution.
package main
import &fmt&
func main() {
ch := make(chan string)
go func() {
for m := range ch {
fmt.Println(&processed:&,m)
ch &- &cmd.1&
ch &- &cmd.2& //won't be processed
Sending to an Closed Channel Causes a Panic
level: beginner
Receiving from a closed channel is safe. The
return value in a receive statement will be set to
indicating that no data was received. If you are receiving from a buffered channel you'll get the buffered data first and once it's empty the
return value will be
Sending data to a closed channel causes a panic. It is a documented behavior, but it's not very intuitive for new Go developers who might expect the send behavior to be similar to the receive behavior.
Using &nil& Channels
level: beginner
Send and receive operations on a
channel block forver. It's a well documented behavior, but it can be a surprise for new Go developers.
This behavior can be used as a way to dynamically enable and disable channels in a
statement.
Methods with Value Receivers Can't Change the Original Value
level: beginner
Method receivers are like regular function arguments. If it's declared to be a value then your function/method gets a copy of your receiver argument. This means making changes to the receiver will not affect the original value unless your receiver is a map or slice variable and you are updating the items in the collection or the fields you are updating in the receiver are pointers.
package main
import &fmt&
type data struct {
key *string
items map[string]bool
func (this *data) pmethod() {
this.num = 7
func (this data) vmethod() {
this.num = 8
*this.key = &v.key&
this.items[&vmethod&] = true
func main() {
key := &key.1&
d := data{1,&key,make(map[string]bool)}
fmt.Printf(&num=%v key=%v items=%v\n&,d.num,*d.key,d.items)
//prints num=1 key=key.1 items=map[]
d.pmethod()
fmt.Printf(&num=%v key=%v items=%v\n&,d.num,*d.key,d.items)
//prints num=7 key=key.1 items=map[]
d.vmethod()
fmt.Printf(&num=%v key=%v items=%v\n&,d.num,*d.key,d.items)
//prints num=7 key=v.key items=map[vmethod:true]
Closing HTTP Response Body
level: intermediate
When you make requests using the standard http library you get a http response variable. If you don't read the response body you still need to close it. Note that you must do it for empty responses too. It's very easy to forget especially for new Go developers.
Some new Go developers do try to close the response body, but they do it in the wrong place.
package main
&net/http&
&io/ioutil&
func main() {
resp, err := http.Get(&https://api.ipify.org?format=json&)
defer resp.Body.Close()//not ok
if err != nil {
fmt.Println(err)
body, err := ioutil.ReadAll(resp.Body)
if err != nil {
fmt.Println(err)
fmt.Println(string(body))
This code works for successful requests, but if the http request fails the
variable might be
, which will cause a runtime panic.
The most common why to close the response body is by using a
call after the http response error check.
package main
&net/http&
&io/ioutil&
func main() {
resp, err := http.Get(&https://api.ipify.org?format=json&)
if err != nil {
fmt.Println(err)
defer resp.Body.Close()//ok, most of the time :-)
body, err := ioutil.ReadAll(resp.Body)
if err != nil {
fmt.Println(err)
fmt.Println(string(body))
Most of the time when your http request fails the
variable will be
variable will be
. However, when you get a redirection failure both variables will be
. This means you can still end up with a leak.
You can fix this leak by adding a call to close
response bodies in the http response error handling block. Another option is to use one
call to close response bodies for all failed and successful requests.
package main
&net/http&
&io/ioutil&
func main() {
resp, err := http.Get(&https://api.ipify.org?format=json&)
if resp != nil {
defer resp.Body.Close()
if err != nil {
fmt.Println(err)
body, err := ioutil.ReadAll(resp.Body)
if err != nil {
fmt.Println(err)
fmt.Println(string(body))
The orignal implementation for
resp.Body.Close()
also reads and discards the remaining response body data. This ensured that the http connection could be reused for another request if the keepalive http connection behavior is enabled. The latest http client behavior is different. Now it's your responsibility to read and discard the remaining response data. If you don't do it the http connection might be closed instead of being reused. This little gotcha is supposed to be documented in Go 1.5.
If reusing the http connection is important for your application you might need to add something like this at the end of your response processing logic:
_, err = io.Copy(ioutil.Discard, resp.Body)
It might be necessary if you don't read the entire response body right away, which might happen if you are processing json http API response with code like this:
json.NewDecoder(resp.Body).Decode(&data)
Closing HTTP Connections
level: intermediate
Some HTTP servers keep network connections open for a while (based on the HTTP 1.1 spec and the server &keep-alive& configurations). By default, the standard http library will close the network connections only when the target HTTP server asks for it. This means your app may run out of sockets/file descriptors under certain conditions.
You can ask the http library to close the connection after your request is done by setting the
field in the request variable to
Another option is to add a
Connection
request header and set it to
. The target HTTP server should respond with a
Connection: close
header too. When the http library sees this response header it will also close the connection.
package main
&net/http&
&io/ioutil&
func main() {
req, err := http.NewRequest(&GET&,&http://golang.org&,nil)
if err != nil {
fmt.Println(err)
req.Close = true
//or do this:
//req.Header.Add(&Connection&, &close&)
resp, err := http.DefaultClient.Do(req)
if resp != nil {
defer resp.Body.Close()
if err != nil {
fmt.Println(err)
body, err := ioutil.ReadAll(resp.Body)
if err != nil {
fmt.Println(err)
fmt.Println(len(string(body)))
You can also disable http connection reuse globally. You'll need to create a custom http transport configuration for it.
package main
&net/http&
&io/ioutil&
func main() {
tr := &http.Transport{DisableKeepAlives: true}
client := &http.Client{Transport: tr}
resp, err := client.Get(&http://golang.org&)
if resp != nil {
defer resp.Body.Close()
if err != nil {
fmt.Println(err)
fmt.Println(resp.StatusCode)
body, err := ioutil.ReadAll(resp.Body)
if err != nil {
fmt.Println(err)
fmt.Println(len(string(body)))
If you send a lot of requests to the same HTTP server it's ok to keep the network connection open. However, if your app sends one or two requests to many different HTTP servers in a short period of time it's a good idea to close the network connections right after your app receives the responses. Increasing the open file limit might be a good idea too. The correct solution depends on your application though.
Comparing Structs, Arrays, Slices, and Maps
level: intermediate
You can use the equality operator,
, to compare struct variables if each structure field can be compared with the equality operator.
package main
import &fmt&
type data struct {
fp float32
complex complex64
str string
events &-chan string
handler interface{}
raw [10]byte
func main() {
v1 := data{}
v2 := data{}
fmt.Println(&v1 == v2:&,v1 == v2) //prints: v1 == v2: true
If any of the struct fields are not comparable then using the equality operator will result in compile time errors. Note that arrays are comparable only if their data items are comparable.
package main
import &fmt&
type data struct {
checks [10]func() bool //not comparable
doit func() bool
//not comparable
m map[string] string
//not comparable
bytes []byte
//not comparable
func main() {
v1 := data{}
v2 := data{}
fmt.Println(&v1 == v2:&,v1 == v2)
Go does provide a number of helper functions to compare variables that can't be compared using the comparison operators.
The most generic solution is to use the
DeepEqual()
function in the reflect package.
package main
type data struct {
checks [10]func() bool //not comparable
doit func() bool
//not comparable
m map[string] string
//not comparable
bytes []byte
//not comparable
func main() {
v1 := data{}
v2 := data{}
fmt.Println(&v1 == v2:&,reflect.DeepEqual(v1,v2)) //prints: v1 == v2: true
m1 := map[string]string{&one&: &a&,&two&: &b&}
m2 := map[string]string{&two&: &b&, &one&: &a&}
fmt.Println(&m1 == m2:&,reflect.DeepEqual(m1, m2)) //prints: m1 == m2: true
s1 := []int{1, 2, 3}
s2 := []int{1, 2, 3}
fmt.Println(&s1 == s2:&,reflect.DeepEqual(s1, s2)) //prints: s1 == s2: true
Aside from being slow (which may or may not be a deal breaker for your application),
DeepEqual()
also has its own gotchas.
package main
func main() {
var b1 []byte = nil
b2 := []byte{}
fmt.Println(&b1 == b2:&,reflect.DeepEqual(b1, b2)) //prints: b1 == b2: false
DeepEqual()
doesn't consider an empty slice to be equal to a &nil& slice. This behavior is different from the behavior you get using the
bytes.Equal()
bytes.Equal()
considers &nil& and empty slices to be equal.
package main
func main() {
var b1 []byte = nil
b2 := []byte{}
fmt.Println(&b1 == b2:&,bytes.Equal(b1, b2)) //prints: b1 == b2: true
DeepEqual()
isn't always perfect comparing slices.
package main
&encoding/json&
func main() {
var str string = &one&
var in interface{} = &one&
fmt.Println(&str == in:&,str == in,reflect.DeepEqual(str, in))
//prints: str == in: true true
v1 := []string{&one&,&two&}
v2 := []interface{}{&one&,&two&}
fmt.Println(&v1 == v2:&,reflect.DeepEqual(v1, v2))
//prints: v1 == v2: false (not ok)
data := map[string]interface{}{
&code&: 200,
&value&: []string{&one&,&two&},
encoded, _ := json.Marshal(data)
var decoded map[string]interface{}
json.Unmarshal(encoded, &decoded)
fmt.Println(&data == decoded:&,reflect.DeepEqual(data, decoded))
//prints: data == decoded: false (not ok)
If your byte slices (or strings) contain text data you might be tempted to use
from the &bytes& and &strings& packages when you need to compare values in a case insensitive manner (before using
bytes.Equal()
). It will work for English text, but it will not work for text in many other languages.
strings.EqualFold()
bytes.EqualFold()
should be used instead.
If your byte slices contain secrets (e.g., cryptographic hashes, tokens, etc.) that need to be validated against user-provided data, don't use
reflect.DeepEqual()
bytes.Equal()
because those functions will make your application vulnerable to
. To avoid leaking the timing information use the functions from the 'crypto/subtle' package (e.g.,
subtle.ConstantTimeCompare()
Recovering From a Panic
level: intermediate
function can be used to catch/intercept a panic. Calling
will do the trick only when it's done in a deferred function.
Incorrect:
package main
import &fmt&
func main() {
recover() //doesn't do anything
panic(&not good&)
recover() //won't be executed :)
fmt.Println(&ok&)
package main
import &fmt&
func main() {
defer func() {
fmt.Println(&recovered:&,recover())
panic(&not good&)
The call to
works only if it's called directly in your deferred function.
package main
import &fmt&
func doRecover() {
fmt.Println(&recovered =&&,recover()) //prints: recovered =& &nil&
func main() {
defer func() {
doRecover() //panic is not recovered
panic(&not good&)
Updating and Referencing Item Values in Slice, Array, and Map &range& Clauses
level: intermediate
The data values generated in the &range& clause are copies of the actual collection elements. They are not references to the original items. This means that updating the values will not change the original data. It also means that taking the address of the values will not give you pointers to the original data.
package main
import &fmt&
func main() {
data := []int{1,2,3}
for _,v := range data {
v *= 10 //original item is not changed
fmt.Println(&data:&,data) //prints data: [1 2 3]
If you need to update the original collection record value use the index operator to access the data.
package main
import &fmt&
func main() {
data := []int{1,2,3}
for i,_ := range data {
data[i] *= 10
fmt.Println(&data:&,data) //prints data: [10 20 30]
If your collection holds pointer values then the rules are slightly different. You still need to use the index operator if you want the original record to point to another value, but you can update the data stored at the target location using the second value in the &for range& clause.
package main
import &fmt&
func main() {
data := []*struct{num int} {{1},{2},{3}}
for _,v := range data {
v.num *= 10
fmt.Println(data[0],data[1],data[2]) //prints &{10} &{20} &{30}
&Hidden& Data in Slices
level: intermediate
When you reslice a slice, the new slice will reference the array of the original slice. If you forget about this behavior it can lead to unexpected memory usage if your application allocates large temporary slices creating new slices from them to refer to small sections of the original data.
package main
import &fmt&
func get() []byte {
raw := make([]byte,10000)
fmt.Println(len(raw),cap(raw),&raw[0]) //prints:
&byte_addr_x&
return raw[:3]
func main() {
data := get()
fmt.Println(len(data),cap(data),&data[0]) //prints: 3 10000 &byte_addr_x&
To avoid this trap make sure to copy the data you need from the temporary slice (instead of reslicing it).
package main
import &fmt&
func get() []byte {
raw := make([]byte,10000)
fmt.Println(len(raw),cap(raw),&raw[0]) //prints:
&byte_addr_x&
res := make([]byte,3)
copy(res,raw[:3])
return res
func main() {
data := get()
fmt.Println(len(data),cap(data),&data[0]) //prints: 3 3 &byte_addr_y&
Slice Data &Corruption&
level: intermediate
Let's say you need to rewrite a path (stored in a slice). You reslice the path to reference each directory modifying the first folder name and then you combine the names to create a new path.
package main
func main() {
path := []byte(&AAAA/BBBBBBBBB&)
sepIndex := bytes.IndexByte(path,'/')
dir1 := path[:sepIndex]
dir2 := path[sepIndex+1:]
fmt.Println(&dir1 =&&,string(dir1)) //prints: dir1 =& AAAA
fmt.Println(&dir2 =&&,string(dir2)) //prints: dir2 =& BBBBBBBBB
dir1 = append(dir1,&suffix&...)
path = bytes.Join([][]byte{dir1,dir2},[]byte{'/'})
fmt.Println(&dir1 =&&,string(dir1)) //prints: dir1 =& AAAAsuffix
fmt.Println(&dir2 =&&,string(dir2)) //prints: dir2 =& uffixBBBB (not ok)
fmt.Println(&new path =&&,string(path))
It didn't work as you expected. Instead of &AAAAsuffix/BBBBBBBBB& you ended up with &AAAAsuffix/uffixBBBB&. It happened because both directory slices referenced the same underlying array data from the original path slice. This means that the original path is also modified. Depending on your application this might be a problem too.
This problem can fixed by allocating new slices and copying the data you need. Another option is to use the full slice expression.
package main
func main() {
path := []byte(&AAAA/BBBBBBBBB&)
sepIndex := bytes.IndexByte(path,'/')
dir1 := path[:sepIndex:sepIndex] //full slice expression
dir2 := path[sepIndex+1:]
fmt.Println(&dir1 =&&,string(dir1)) //prints: dir1 =& AAAA
fmt.Println(&dir2 =&&,string(dir2)) //prints: dir2 =& BBBBBBBBB
dir1 = append(dir1,&suffix&...)
path = bytes.Join([][]byte{dir1,dir2},[]byte{'/'})
fmt.Println(&dir1 =&&,string(dir1)) //prints: dir1 =& AAAAsuffix
fmt.Println(&dir2 =&&,string(dir2)) //prints: dir2 =& BBBBBBBBB (ok now)
fmt.Println(&new path =&&,string(path))
The extra parameter in the full slice expression controls the capacity for the new slice. Now appending to that slice will trigger a new buffer allocation instead of overwriting the data in the second slice.
&Stale& Slices
level: intermediate
Multiple slices can reference the same data. This can happen when you create a new slice from an existing slice, for example. If your application relies on this behavior to function properly then you'll need to worry about &stale& slices.
At some point adding data to one of the slices will result in a new array allocation when the original array can't hold any more new data. Now other slices will point to the old array (with old data).
import &fmt&
func main() {
s1 := []int{1,2,3}
fmt.Println(len(s1),cap(s1),s1) //prints 3 3 [1 2 3]
s2 := s1[1:]
fmt.Println(len(s2),cap(s2),s2) //prints 2 2 [2 3]
for i := range s2 { s2[i] += 20 }
//still referencing the same array
fmt.Println(s1) //prints [1 22 23]
fmt.Println(s2) //prints [22 23]
s2 = append(s2,4)
for i := range s2 { s2[i] += 10 }
//s1 is now &stale&
fmt.Println(s1) //prints [1 22 23]
fmt.Println(s2) //prints [32 33 14]
Type Declarations and Methods
level: intermediate
When you create a type declaration by defining a new type from an existing (non-interface) type, you don't inherit the methods defined for that existing type.
package main
import &sync&
type myMutex sync.Mutex
func main() {
var mtx myMutex
mtx.Lock() //error
mtx.Unlock() //error
Compile Errors:
/tmp/sandbox/main.go:9: mtx.Lock undefined (type myMutex has no field or method Lock) /tmp/sandbox/main.go:10: mtx.Unlock undefined (type myMutex has no field or method Unlock)
If you do need the methods from the original type you can define a new struct type embedding the original type as an anonymous field.
package main
import &sync&
type myLocker struct {
sync.Mutex
func main() {
var lock myLocker
lock.Lock() //ok
lock.Unlock() //ok
Interface type declarations also retain their method sets.
package main
import &sync&
type myLocker sync.Locker
func main() {
var lock myLocker = new(sync.Mutex)
lock.Lock() //ok
lock.Unlock() //ok
Breaking Out of &for switch& and &for select& Code Blocks
level: intermediate
A &break& statement without a label only gets you out of the inner switch/select block. If using a &return& statement is not an option then defining a label for the outer loop is the next best thing.
package main
import &fmt&
func main() {
case true:
fmt.Println(&breaking out...&)
break loop
fmt.Println(&out!&)
A &goto& statement will do the trick too...
Iteration Variables and Closures in &for& Statements
level: intermediate
This is the most common gotcha in Go. The iteration variables in
statements are reused in each iteration. This means that each closure (aka function literal) created in your
loop will reference the same variable (and they'll get that variable's value at the time those goroutines start executing).
Incorrect:
package main
func main() {
data := []string{&one&,&two&,&three&}
for _,v := range data {
go func() {
fmt.Println(v)
time.Sleep(3 * time.Second)
//goroutines print: three, three, three
The easiest solution (that doesn't require any changes to the goroutine) is to save the current iteration variable value in a local variable inside the
loop block.
package main
func main() {
data := []string{&one&,&two&,&three&}
for _,v := range data {
vcopy := v //
go func() {
fmt.Println(vcopy)
time.Sleep(3 * time.Second)
//goroutines print: one, two, three
Another solution is to pass the current iteration variable as a parameter to the anonymous goroutine.
package main
func main() {
data := []string{&one&,&two&,&three&}
for _,v := range data {
go func(in string) {
fmt.Println(in)
time.Sleep(3 * time.Second)
//goroutines print: one, two, three
Here's a slightly more complicated version of the trap.
Incorrect:
package main
type field struct {
name string
func (p *field) print() {
fmt.Println(p.name)
func main() {
data := []field{{&one&},{&two&},{&three&}}
for _,v := range data {
go v.print()
time.Sleep(3 * time.Second)
//goroutines print: three, three, three
package main
type field struct {
name string
func (p *field) print() {
fmt.Println(p.name)
func main() {
data := []field{{&one&},{&two&},{&three&}}
for _,v := range data {
go v.print()
time.Sleep(3 * time.Second)
//goroutines print: one, two, three
What do you think you'll see when you run this code (and why)?
package main
type field struct {
name string
func (p *field) print() {
fmt.Println(p.name)
func main() {
data := []*field{{&one&},{&two&},{&three&}}
for _,v := range data {
go v.print()
time.Sleep(3 * time.Second)
Deferred Function Call Argument Evaluation
level: intermediate
Arguments for a deferred function call are evaluated when the
statement is evaluated (not when the function is actually executing).
package main
import &fmt&
func main() {
var i int = 1
defer fmt.Println(&result =&&,func() int { return i * 2 }())
//prints: result =& 2 (not ok if you expected 4)
Deferred Function Call Execution
level: intermediate
The deferred calls are executed at the end of the containing function and not at the end of the containing code block. It's an easy mistake to make for new Go developers confusing the deferred code execution rules with the variable scoping rules. It can become a problem if you have a long running piece of code with a
loop that tries to
resource cleanup calls in each iteration.
Failed Type Assertions
level: intermediate
Failed type assertions return the &zero value& for the target type used in the assertion statement. This can lead to unexpected behavior when it's mixed with variable shadowing.
Incorrect:
package main
import &fmt&
func main() {
var data interface{} = &great&
if data, ok := data.(int); ok {
fmt.Println(&[is an int] value =&&,data)
fmt.Println(&[not an int] value =&&,data)
//prints: [not an int] value =& 0 (not &great&)
package main
import &fmt&
func main() {
var data interface{} = &great&
if res, ok := data.(int); ok {
fmt.Println(&[is an int] value =&&,res)
fmt.Println(&[not an int] value =&&,data)
//prints: [not an int] value =& great (as expected)
Blocked Goroutines and Resource Leaks
level: intermediate
Rob Pike talked about a number of fundamental concurrency patterns in his
presentation at Google I/O in 2012. Fetching the first result from a number of targets is one of them.
func First(query string, replicas ...Search) Result {
c := make(chan Result)
searchReplica := func(i int) { c &- replicas[i](query) }
for i := range replicas {
go searchReplica(i)
return &-c
The function starts a goroutines for each search replica. Each goroutine sends its search result to the result channel. The first value from the result channel is returned.
What about the results from the other goroutines? What about the goroutines themselves?
The result channel in the
function is unbuffered. This means that only the first goroutine returns. All other goroutines are stuck trying to send their results. This means if you have more than one replica each call will leak resources.
To avoid the leaks you need to make sure all goroutines exit. One potential solution is to use a buffered result channel big enough to hold all results.
func First(query string, replicas ...Search) Result {
c := make(chan Result,len(replicas))
searchReplica := func(i int) { c &- replicas[i](query) }
for i := range replicas {
go searchReplica(i)
return &-c
Another potential solution is to use a
statement with a
case and a buffered result channel that can hold one value. The
case ensures that the goroutines don't get stuck even when the result channel can't receive messages.
func First(query string, replicas ...Search) Result {
c := make(chan Result,1)
searchReplica := func(i int) {
case c &- replicas[i](query):
for i := range replicas {
go searchReplica(i)
return &-c
You can also use a special cancellation channel to interrupt the workers.
func First(query string, replicas ...Search) Result {
c := make(chan Result)
done := make(chan struct{})
searchReplica := func(i int) {
case c &- replicas[i](query):
case &- done:
for i := range replicas {
go searchReplica(i)
result := &-c
close(done)
return result
Why did the presentation contain these bugs? Rob Pike simply didn't want to comlicate the slides. It makes sense, but it can be a problem for new Go developers who would use the code as is without thinking that it might have problems.
Using Pointer Receiver Methods On Value Instances
level: advanced
It's OK to call a pointer receiver method on a value as long as the value is addressable. In other words, you don't need to have a value receiver version of the method in some cases.
Not every variable is addressable though. Map elements are not addressable. Variables referenced through interfaces are also not addressable.
package main
import &fmt&
type data struct {
name string
func (p *data) print() {
fmt.Println(&name:&,p.name)
type printer interface {
func main() {
d1 := data{&one&}
d1.print() //ok
var in printer = data{&two&} //error
in.print()
m := map[string]data {&x&:data{&three&}}
m[&x&].print() //error
Compile Errors:
/tmp/sandbox/main.go:21: cannot use data literal (type data) as type printer in assignment: data does not implement printer (print method has pointer receiver)
/tmp/sandbox/main.go:25: cannot call pointer method on m[&x&] /tmp/sandbox/main.go:25: cannot take the address of m[&x&]
Updating Map Value Fields
level: advanced
If you have a map of struct values you can't update individual struct fields.
package main
type data struct {
name string
func main() {
m := map[string]data {&x&:{&one&}}
m[&x&].name = &two& //error
Compile Error:
/tmp/sandbox/main.go:9: cannot assign to m[&x&].name
It doesn't work because map elements are not addressable.
What can be extra confusing for new Go devs is the fact that slice elements are addressable.
package main
import &fmt&
type data struct {
name string
func main() {
s := []data {{&one&}}
s[0].name = &two& //ok
fmt.Println(s) //prints: [{two}]
Note that a while ago it was possible to update map element fields in one of the Go compilers (gccgo), but that behavior was quickly fixed :-) It was also considered as a potential feature for Go 1.3. It wasn't important enough to support at that point in time, so it's still on the todo list.
The first work around is to use a temporary variable.
package main
import &fmt&
type data struct {
name string
func main() {
m := map[string]data {&x&:{&one&}}
r := m[&x&]
r.name = &two&
m[&x&] = r
fmt.Printf(&%v&,m) //prints: map[x:{two}]
Another workaround is to use a map of pointers.
package main
import &fmt&
type data struct {
name string
func main() {
m := map[string]*data {&x&:{&one&}}
m[&x&].name = &two& //ok
fmt.Println(m[&x&]) //prints: &{two}
By the way, what happens when you run this code?
package main
type data struct {
name string
func main() {
m := map[string]*data {&x&:{&one&}}
m[&z&].name = &what?& //???
&nil& Interfaces and &nil& Interfaces Values
level: advanced
This is the second most common gotcha in Go because interfaces are not pointers even though they may look like pointers. Interface variables will be &nil& only when their type and value fields are &nil&.
The interface type and value fields are populated based on the type and value of the variable used to create the corresponding interface variable. This can lead to unexpected behavior when you are trying to check if an interface variable equals to &nil&.
package main
import &fmt&
func main() {
var data *byte
var in interface{}
fmt.Println(data,data == nil) //prints: &nil& true
fmt.Println(in,in == nil)
//prints: &nil& true
fmt.Println(in,in == nil)
//prints: &nil& false
//'data' is 'nil', but 'in' is not 'nil'
Watch out for this trap when you have a function that returns interfaces.
Incorrect:
package main
import &fmt&
func main() {
doit := func(arg int) interface{} {
var result *struct{} = nil
if(arg & 0) {
result = &struct{}{}
return result
if res := doit(-1); res != nil {
fmt.Println(&good result:&,res) //prints: good result: &nil&
//'res' is not 'nil', but its value is 'nil'
package main
import &fmt&
func main() {
doit := func(arg int) interface{} {
var result *struct{} = nil
if(arg & 0) {
result = &struct{}{}
return nil //return an explicit 'nil'
return result
if res := doit(-1); res != nil {
fmt.Println(&good result:&,res)
fmt.Println(&bad result (res is nil)&) //here as expected
Stack and Heap Variables
level: advanced
You don't always know if your variable is allocated on the stack or heap. In C++ creating variables using the
operator always means that you have a heap variable. In Go the compiler decides where the variable will be allocated even if the
functions are used. The compiler picks the location to store the variable based on its size and the result of &escape analysis&. This also means that it's ok to return references to local variables, which is not ok in other languages like C or C++.
If you need to know where your variables are allocated pass the &-m& gc flag to &go build& or &go run& (e.g.,
go run -gcflags -m app.go
GOMAXPROCS, Concurrency, and Parallelism
level: advanced
By default, Go uses only one execution context / OS thread (in the current version). This number can be increased by setting
GOMAXPROCS
There's a common misconception that
GOMAXPROCS
represents the number of CPUs Go will use to run goroutines. The
runtime.GOMAXPROCS()
function documentation adds more to the confusion. The
GOMAXPROCS
variable description (
) does a better job talking about OS threads.
You can set
GOMAXPROCS
to more than the number of your CPUs. The max value for
GOMAXPROCS
package main
func main() {
fmt.Println(runtime.GOMAXPROCS(-1)) //prints: 1
fmt.Println(runtime.NumCPU())
//prints: 1 (on play.golang.org)
runtime.GOMAXPROCS(20)
fmt.Println(runtime.GOMAXPROCS(-1)) //prints: 20
runtime.GOMAXPROCS(300)
fmt.Println(runtime.GOMAXPROCS(-1)) //prints: 256
Read and Write Operation Reordering
level: advanced
Go may reorder some operations, but it ensures that the overall behavior in the goroutine where it happens doesn't change. However, it doesn't guarantee the order of execution across multiple goroutines.
package main
var _ = runtime.GOMAXPROCS(3)
var a, b int
func u1() {
func u2() {
func p() {
println(a)
println(b)
func main() {
time.Sleep(1 * time.Second)
If you run this code a few times you might see these
variable combinations:
The most interesting combination for
is &02&. It shows that
was updated before
If you need to preserve the order of read and write operations across multiple goroutines you'll need to use channels or the appropriate constructs from the &sync& package.
Preemptive Scheduling
level: advanced
It's possible to have a rogue goroutine that prevents other goroutines from running. It can happen if you have a
loop that doesn't allow the scheduler to run.
package main
import &fmt&
func main() {
done := false
go func(){
done = true
for !done {
fmt.Println(&done!&)
loop doesn't have to be empty. It'll be a problem as long as it contains code that doesn't trigger the scheduler execution.
The scheduler will run after GC, &go& statements, blocking channel operations, blocking system calls, and lock operations. It may also run when a non-inlined function is called.
package main
import &fmt&
func main() {
done := false
go func(){
done = true
for !done {
fmt.Println(&not done!&) //not inlined
fmt.Println(&done!&)
To find out if the function you call in the
loop is inlined pass the &-m& gc flag to &go build& or &go run& (e.g.,
go build -gcflags -m
Another option is to invoke the scheduler explicitly. You can do it with the
function from the &runtime& package.
package main
func main() {
done := false
go func(){
done = true
for !done {
runtime.Gosched()
fmt.Println(&done!&)
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