[go]从os.Stdin探究文件类源码

咋一看go的标准输入输出函数有一箩筐. 细究了一下.

- 从标准输入获取输入
fmt.Scan   以空白(空格或换行)分割,值满后无结束
fmt.Scanln 以空格作为分割,遇到换行结束
fmt.Scanf  以空格作为分隔符,回车结束

- 从字符串获取输入
fmt.Sscan
fmt.Sscanln
fmt.Sscanf

- 从文件获取输入
fmt.Fscan
fmt.Fscanln
fmt.Fscanf



- 输出
fmt.Print
fmt.Println
fmt.Printf

- 格式化输出并返回str
fmt.Sprint
fmt.Sprintln
fmt.Sprintf

fmt.Scan使用参考

Go语言标准库之fmt.Scan - aaronthon - 博客园

os.Stdin使用

os.Stdin.Write
os.Stdin.WriteSring

Stdin  = NewFile(uintptr(syscall.Stdin), "/dev/stdin")

func NewFile(fd uintptr, name string) *File {
        h := syscall.Handle(fd)
        if h == syscall.InvalidHandle {
                return nil
        }
        return newFile(h, name, "file")
}

func newFile(h syscall.Handle, name string, kind string) *File {
        if kind == "file" {
                var m uint32
                if syscall.GetConsoleMode(h, &m) == nil {
                        kind = "console"
                }
                if t, err := syscall.GetFileType(h); err == nil && t == syscall.FILE_TYPE_PIPE {
                        kind = "pipe"
                }
        }

        f := &File{&file{
                pfd: poll.FD{
                        Sysfd:         h,
                        IsStream:      true,
                        ZeroReadIsEOF: true,
                },
                name: name,
        }}
        runtime.SetFinalizer(f.file, (*file).close)

        // Ignore initialization errors.
        // Assume any problems will show up in later I/O.
        f.pfd.Init(kind, false)

        return f
}

type File struct {
        *file // os specific
}

type file struct {
        pfd        poll.FD
        name       string
        dirinfo    *dirInfo // nil unless directory being read
        appendMode bool     // whether file is opened for appending
}

func (f *File) WriteString(s string) (n int, err error) {
        return f.Write([]byte(s))
}

简化如下

type File struct{
   *file
}


//绑定方法
func(f *File)Write(){

}
func(f *File)WriteString(){

}


//构造方法
func newFile(){
    return &File{}
}

os.Stdin = newFile(xxx)