捕获函数调用的标准输出

Question

实现这个测试助手函数有什么重大缺陷吗？

我知道，首选的方法(重构)只会使要测试的函数返回其值并将打印委托给另一个部分，但有时必须测试其中的内容。

import (
    "io"
    "os"
)

// captureStdout calls a function f and returns its stdout side-effect as string
func captureStdout(f func()) string {
    // return to original state afterwards
    // note: defer evaluates (and saves) function ARGUMENT values at definition
    // time, so the original value of os.Stdout is preserved before it is changed
    // further into this function.
    defer func(orig *os.File) {
        os.Stdout = orig
    }(os.Stdout)

    r, w, _ := os.Pipe()
    os.Stdout = w
    f()
    w.Close()
    out, _ := io.ReadAll(r)

    return string(out)
}

动机全示例

main.go

package main

import "fmt"

func main() {
    fmt.Println("Hello, World!")
}

main_test.go

package main

import (
    "io"
    "os"
    "testing"
)

func Test_main(t *testing.T) {
    want := "Hello, World!\n"
    got := captureStdout(main)
    if got != want {
        t.Errorf("main() = %v, want %v", got, want)
    }
}

// captureStdout calls a function f and returns its stdout side-effect as string
func captureStdout(f func()) string {
    // return to original state afterwards
    // note: defer evaluates (and saves) function ARGUMENT values at definition
    // time, so the original value of os.Stdout is preserved before it is
    // changed further into this function.
    defer func(orig *os.File) {
        os.Stdout = orig
    }(os.Stdout)

    r, w, _ := os.Pipe()
    os.Stdout = w
    f()
    w.Close()
    out, _ := io.ReadAll(r)

    return string(out)
}

Answer 1

r, w, \_ := os.Pipe()

为什么我们忽略错误返回？我们真的相信没有办法让管道失效吗？即使在一堆打开的()S没有关闭()？

out, \_ := io.ReadAll(r)

更令人费解的是。阅读错误永远不会发生吗？

对不起，我不能建议以当前状态传送此代码。