Why does Golang handle closures differently in goroutines?

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我寻月下人不归
我寻月下人不归 2020-11-30 10:51

Consider the following Golang code (also on the Go Playground):

package main

import \"fmt\"
import \"time\"

func main() {
    for _, s := range []string{\"         


        
2条回答
  •  旧巷少年郎
    2020-11-30 11:21

    Tip: You can use the "get address operator" & to confirm whether or not variables are the same.

    Let's slightly modify your program to help our understanding.

    package main
    
    import "fmt"
    import "time"
    
    func main() {
        for _, s := range []string{"foo", "bar"} {
            x := s
            fmt.Println("  &s =", &s, "\t&x =", &x)
            func() {
                fmt.Println("-", "&s =", &s, "\t&x =", &x)
                fmt.Println("s =", s, ", x =", x)
            }()
        }
    
        fmt.Println("\n\n")
    
        for _, s := range []string{"foo", "bar"} {
            x := s
            fmt.Println("  &s =", &s, "\t&x =", &x)
            go func() {
                fmt.Println("-", "&s =", &s, "\t&x =", &x)
                fmt.Println("s =", s, ", x =", x)
            }()
        }
        time.Sleep(time.Second)
    }
    

    The output is:

      &s = 0x1040a120   &x = 0x1040a128
    - &s = 0x1040a120   &x = 0x1040a128
    s = foo , x = foo
      &s = 0x1040a120   &x = 0x1040a180
    - &s = 0x1040a120   &x = 0x1040a180
    s = bar , x = bar
    
    
    
      &s = 0x1040a1d8   &x = 0x1040a1e0
      &s = 0x1040a1d8   &x = 0x1040a1f8
    - &s = 0x1040a1d8   &x = 0x1040a1e0
    s = bar , x = foo
    - &s = 0x1040a1d8   &x = 0x1040a1f8
    s = bar , x = bar
    

    Key points:

    • The variable s in each iteration of the loop is the same variable.
    • The local variable x in each iteration of the loop are different variables, they just happen to have the same name x
    • In the first for loop, the func () {} () part got executed in each iteration and the loop only continue to its next iteration after func () {} () completed.
    • In the second for loop (goroutine version), the go func () {} () statement itself completed instantaneously. When the statements in the func body got executed is determined by the Go scheduler. But when they (the statements in the func body) starts to execute, the for loop already completed! And the variable s is the last element in the slice which is bar. That's why we got two "bar"s in the second for loop output.

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