Go - 延迟函数的不一致评估

Question

我正在尝试使用 Go 并且看到一些带有延迟函数的意外行为。考虑以下将全局变量增加给定数量的程序。package mainimport "fmt"var z = 1func main() {    defer increaseZ(10)    defer fmt.Println("z =", increaseZ(20), "Deferred Value 1")    defer fmt.Println("z =", increaseZ(30), "Deferred Value 2")    fmt.Println("z =", z, "Main Value")}func increaseZ(y int) int {    z += y    println("z =", z, "Inside Increase Function")    return z}当在 go playground 中运行时，输出：z = 21 Inside Increase Functionz = 51 Inside Increase Functionz = 61 Inside Increase Functionz = 51 Main Valuez = 51 Deferred Value 2z = 21 Deferred Value 1如果我切换延迟函数的顺序，它还有另一个效果：defer fmt.Println("z =", increaseZ(20), "Deferred Value 1")defer fmt.Println("z =", increaseZ(30), "Deferred Value 2")defer increaseZ(10)输出：z = 21 Inside Increase Functionz = 51 Inside Increase Functionz = 51 Main Valuez = 61 Inside Increase Functionz = 51 Deferred Value 2z = 21 Deferred Value 1Go 文档指出：延迟调用的参数会立即计算，但直到周围的函数返回时才会执行函数调用。因此，正在评估的参数可以解释为什么返回的 Main Value 是 51 而不是 61，因为 fmt.Println 语句将increaseZ作为参数，但defer increaseZ(10)在 main 函数返回之前不会被调用。但是，这并不能解释为什么在第一个示例中increaseZ(10)在 main 完成之前输出，而在第二个示例中在 main 完成之后输出。如果有人能帮助我了解这里发生的事情，我将不胜感激，因为这看起来像是难以诊断错误的沃土。

Answer 1

您的打印目的地不一致。

stdout: fmt.Println

stderr: println

写入相同的打印目的地。

package main

import "fmt"

var z = 1

func main() {

    defer increaseZ(10)

    defer fmt.Println("z =", increaseZ(20), "Deferred Value 1")

    defer fmt.Println("z =", increaseZ(30), "Deferred Value 2")

    fmt.Println("z =", z, "Main Value")

}

func increaseZ(y int) int {

    z += y

    fmt.Println("z =", z, "Inside Increase Function")

    return z

}

输出：

z = 21 Inside Increase Function

z = 51 Inside Increase Function

z = 51 Main Value

z = 51 Deferred Value 2

z = 21 Deferred Value 1

z = 61 Inside Increase Function

或者，

package main

import (

    "fmt"

    "os"

)

var z = 1

func main() {

    defer increaseZ(10)

    defer fmt.Fprintln(os.Stderr, "z =", increaseZ(20), "Deferred Value 1")

    defer fmt.Fprintln(os.Stderr, "z =", increaseZ(30), "Deferred Value 2")

    fmt.Fprintln(os.Stderr, "z =", z, "Main Value")

}

func increaseZ(y int) int {

    z += y

    println("z =", z, "Inside Increase Function")

    return z

}

输出：

z = 21 Inside Increase Function

z = 51 Inside Increase Function

z = 51 Main Value

z = 51 Deferred Value 2

z = 21 Deferred Value 1

z = 61 Inside Increase Function

Answer 2

我怀疑这是 Go 游乐场的错误。当我在我的机器上编译并运行这个程序时，它会产生预期的输出。一个bug报告关于这个问题已经提交。