CODEWITHSUNDEEP
go
hol
hol

Reputation: 33

Read from io.PipeReader does not match write to io.PipeWriter

Running this program:

on play.golang.org

package main

import (
    "bytes"
    "fmt"
    "io"
)

func main() {
    r, w := io.Pipe()

    go func() {
        defer w.Close()
        in := 'a' // single quote
        fmt.Println("writing: ", in)
        fmt.Fprint(w, in)
    }()

    defer r.Close()

    buf := new(bytes.Buffer)
    buf.ReadFrom(r)

    fmt.Print("read: ", buf.Bytes())

}

This outputs:

writing:  97
read: [57 55]

When I do in := "a" or fmt.Fprint(w, string(r)), it works as expected (read: [97]).

Is 97 equal to [57, 55] somehow?

Upvotes: 1

Views: 476

Answers (1)

Darshan Rivka Whittle
Darshan Rivka Whittle

Reputation: 34031

Indeed, the two bytes 57 and 55 are the same as the two characters in the string "97". If you change your last line to use String() rather than Bytes(), you'll get the string:

fmt.Print("read: ", buf.String())

When you use Bytes(), you're getting the byte value of "9", which is 57, followed by that of "7", which is 55.

As pointed out by Peter in the comments, you might not be understanding what fmt.Fprint is doing—it's converting the rune 'a' to the string "97" (this is the same as how fmt.Println is giving you "writing: 97"). If you don't want to do that conversion when you write to the pipe, then don't use fmt.Printf. As Peter suggests, w.Write is a good option. For example, w.Write([]byte{byte(in)}) does what you seem to want, and allows the final line to work the way you want unchanged. Another option would be fmt.Fprintf(w, "%c", in).

Upvotes: 1

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