duannan3959 2018-11-15 17:57
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Golang闭包不存储捕获的变量?

In this block of code below, I expect the output to be 0, 1, 2, 3,..., 9. However, only the output 10, 10,...,10 is produced.

package main

import "fmt"

func adder() []func() {
    out := []func(){}
    for i := 0; i < 10; i++ {
        out = append(out, func() { fmt.Println(i) })
    }

    return out
}

func main() {
    out := adder()
    for i := 0; i < 10; i++ {
        out[i]()
    }
}

My mental model suggests that a new function is generated and stored every time in the for-loop, but it doesn't seem to be the case. Is there a solution to generating new functions without modifying the signature (no parameters or global variables) but still keeping the current value of i?

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1条回答 默认 最新

  • dongtan8979 2018-11-15 18:00
    关注

    A new function is created every iteration, but they all close over the same variable, i. That variable is updated with every iteration until the last, so when you go through and execute all those functions, they all refer to the same i, with the same value, 10.

    If you don't want that behavior, you must not close over the loop variable:

    for i := 0; i < 10; i++ {
        temp := i
        out = append(out, func() { fmt.Println(temp) })
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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