Golang Go语言中为啥解压zip文件要这么多代码

这简直不能忍啊

C#版

ZipFile.ExtractToDirectory("source", "target");

Golang 版

func Unzip(src, dest string) error {
    r, err := zip.OpenReader(src)
    if err != nil {
        return err
    }
    defer func() {
        if err := r.Close(); err != nil {
            panic(err)
        }
    }()
os.MkdirAll(dest, 0755)

// Closure to address file descriptors issue with all the deferred .Close() methods
extractAndWriteFile := func(f *zip.File) error {
    rc, err := f.Open()
    if err != nil {
        return err
    }
    defer func() {
        if err := rc.Close(); err != nil {
            panic(err)
        }
    }()

    path := filepath.Join(dest, f.Name)

    // Check for ZipSlip (Directory traversal)
    if !strings.HasPrefix(path, filepath.Clean(dest) + string(os.PathSeparator)) {
        return fmt.Errorf("illegal file path: %s", path)
    }

    if f.FileInfo().IsDir() {
        os.MkdirAll(path, f.Mode())
    } else {
        os.MkdirAll(filepath.Dir(path), f.Mode())
        f, err := os.OpenFile(path, os.O_WRONLY|os.O_CREATE|os.O_TRUNC, f.Mode())
        if err != nil {
            return err
        }
        defer func() {
            if err := f.Close(); err != nil {
                panic(err)
            }
        }()

        _, err = io.Copy(f, rc)
        if err != nil {
            return err
        }
    }
    return nil
}

for _, f := range r.File {
    err := extractAndWriteFile(f)
    if err != nil {
        return err
    }
}

return nil
}