go - 在另一个函数中更改golang slice

Question

我有一个 slice ，如果我直接在主函数中从中删除一个元素，则 slice 的长度将减少一。但是在另一个函数中执行remove并在主函数中调用它， slice 的长度仍然保持原点。谁能为我解释？谢谢!

package main

import "fmt"

func main() {
    a := []int{1, 2, 3, 4}
    i := 0

    //copy(a[i:], a[i+1:])
    //a[len(a)-1] = 0
    //a = a[:len(a)-1]
    //fmt.Println(a)    //outputs: [2 3 4], this is correct

    f(a, i)
    fmt.Println(a) //outputs: [2 3 4 0], this is wrong!
}

func f(a []int, i int) {
    copy(a[i:], a[i+1:])
    a[len(a)-1] = 0
    a = a[:len(a)-1]
    fmt.Println(a) //outputs: [2 3 4], here still correct
}

Go Playground Link

Answer 1

slice 按值传递，因此在函数f中对其进行更改不会在main函数中对其进行更改。您可以像这样通过指针传递:

package main

import "fmt"

func main() {
    a := []int{1, 2, 3, 4}
    i := 0

    f(&a, i)
    fmt.Println(a)    //outputs: [2 3 4], correct
}

func f(a *[]int, i int) {
    b := *a
    copy(b[i:], b[i+1:])
    // The following line seems pointless, but ok...
    b[len(b)-1] = 0
    b = b[:len(b)-1]
    fmt.Println(b)    //outputs: [2 3 4], here still correct
    *a = b
}

Go Playground

正如@zerkms在评论中所建议的那样，您还可以返回新的 slice ，从而避免使用指针:

package main

import "fmt"

func main() {
    a := []int{1, 2, 3, 4}
    i := 0

    a = f(a, i)
    fmt.Println(a)
}

func f(a []int, i int) []int {
    copy(a[i:], a[i+1:])
    // The following line seems pointless, but ok...
    a[len(a)-1] = 0
    a = a[:len(a)-1]
    fmt.Println(a)     //outputs: [2 3 4], here still correct
    return a
}