使用值接收器附加到具有足够容量的切片 [英] Appending to a slice with enough capacity using value receiver

问题描述

有人可以帮助我了解这里发生的事情吗?

can someone help me understand what happens here?

package main

import (
    "fmt"
)

func appendString(slice []string, newString string) {
    slice = append(slice, newString)
}

func main() {
    slice := make([]string, 0, 1)
    appendString(slice, "a")
    fmt.Println(slice)
}

我了解切片标头以及使用指针接收器的需要.但是在这里，由于基础数组具有足够的容量，因此我希望追加仍可以正常工作(只需将新值添加到基础数组中，并且原始的[copied]标头可以按预期工作)

I know about the slice header and the need to use a pointer receiver. But here, as the underlying array has enough capacity I would expect append to work anyways (just adding the new value to the underlying array and the original [copied] header working as expected)

我的假设有什么问题?

推荐答案

让我们添加最终的打印语句以查看结果:

Let's add a final print statement to see the result:

slice := make([]string, 0, 1)
fmt.Println(cap(slice))
appendString(slice, "a")
fmt.Println(slice)

输出将是(在进入游乐场上尝试):

And the output will be (try it on the Go Playground):

1
[]

这是正确的.可以期望输出为:

Which is correct. One could expect the output to be:

1
[a]

之所以不是这样，是因为即使不分配新的后备数组，main()内的slice变量中的slice头也不会不变，它将仍然按住length = 0.仅存储在appendString()(参数)内部的slice局部变量中的slice头，但是此变量独立于main的slice.

The reason why this is not the case is because even though a new backing array will not be allocated, the slice header in the slice variable inside main() is not changed, it will still hold length = 0. Only the slice header stored in the slice local variable inside appendString() (the parameter) is changed, but this variable is independent from main's slice.

如果要重新分割main的slice，您将看到支持数组确实包含新字符串:

If you were to reslice main's slice, you will see that the backing array does contain the new string:

slice := make([]string, 0, 1)
fmt.Println(cap(slice))
appendString(slice, "a")
fmt.Println(slice)

slice = slice[:1]
fmt.Println(slice)

现在将输出(在进入游乐场上尝试):

Now output will be (try it on the Go Playground):

1
[]
[a]

这就是为什么内置 append() 必须返回新切片的原因:即使不需要新的后备数组，如果追加了0个以上的元素，也必须更改(增加)切片头(包含长度).

This is why the builtin append() has to return the new slice: because even if no new backing array is needed, the slice header (which contains the length) will have to be changed (increased) if more than 0 elements are appended.

这就是为什么appendString()还应该返回新切片的原因:

This is why appendString() should also return the new slice:

func appendString(slice []string, newString string) []string {
    slice = append(slice, newString)
    return slice
}

或更短:

func appendString(slice []string, newString string) []string {
    return append(slice, newString)
}

您必须重新分配使用位置:

Which you have to reassign where you use it:

slice := make([]string, 0, 1)
fmt.Println(cap(slice))
slice = appendString(slice, "a")
fmt.Println(slice)

然后立即获得预期的结果(在进入游乐场上尝试) :

And then you get the expected outcome right away (try it on the Go Playground):

1
[a]

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