Golang中的字符串解码

Question

In Ruby, I'm using the string.unpack method to decode some data. I'm migrating this code into a Golang app, however I can't find a corresponding method in the standard library?

The ruby code is: str.unpack('>LLLff')

What is the best way to implement this in Go?

---

Example data:

]\x00\x00@\x00\x8C\xD2\x00\x00\x00\x00\x00\x00\x00\x00h.\x9FAjK\x8FG\xC0T\x87UX\xA1CH\xB7\x80\x82}\xBB\xF9\x8F\xBAcqF{\xE31/Zx\xC0\x04\xFF\b\xA0\xBD\a\x9E\x0F\x14c+\b\xD3\xE8\xE1\xC2\x1C7\xBCmd/j\xC1H\x191~\x91F\b5-\x8A\x9F(\\\xF1\x16/\x0F\xEA\xD9\x94rV\xEC\xBB\xCB\x12qJ\x05\b\x04 {R\xB8\xA4\xA9\xD0)v\x97<\xCA\x92\xB6U\b)\x1D\x1D\xF1\x00\xE4U\x8DB\x16/v\xEC[bLTW\xB2\xFB\x0F\x91\x05\xAA\x9C\xBD\x8E\x1A\xFB\xDFUJ\xB8\xDE\xF5m\xC8Z\xC1\v\\|;\xB6\xE6\xA6\x13o\x91\xE1\xBA\x9An\x8B\x01W\xB9\"\x12+\x9B[\xC7\xCD\xC2O+\f^|\xE1G\x86.%m\xB2x\xF1j\e\"\xE41Bu_\x16J\xE4M\x7F\xC5U[\xB9\xD0\xC7\xECu\xE53\xB4)\xD5\x80|\xB7.\xD2\f\xF3\x16\x8B\xE3\x13\x81\x02:5\x88\x0F

Should map to: [1073741917, 13798400, 0, 2.4178516392292583e+24, 5.851859700165677e+25]

Answer 1

Try this:

package main

import (
    "bytes"
    "encoding/binary"
    "log"
)

type UnpackedThing struct {
    First  uint32
    Second uint32
    Third  uint32

    Fourth float32
    Fifth  float32
}

func main() {
    input := bytes.NewBufferString("]\x00\x00@\x00\x8C\xD2\x00\x00\x00\x00\x00\x00\x00\x00h.\x9FAjK\x8FG\xC0T\x87UX\xA1CH\xB7\x80\x82}\xBB\xF9\x8F\xBAcqF{\xE31/Zx\xC0\x04\xFF\b\xA0\xBD\a\x9E\x0F\x14c+\b\xD3\xE8\xE1\xC2\x1C7\xBCmd/j\xC1H\x191~\x91F\b5-\x8A\x9F(\\\xF1\x16/\x0F\xEA\xD9\x94rV\xEC\xBB\xCB\x12qJ\x05\b\x04 {R\xB8\xA4\xA9\xD0)v\x97<\xCA\x92\xB6U\b)\x1D\x1D\xF1\x00\xE4U\x8DB\x16/v\xEC[bLTW\xB2\xFB\x0F\x91\x05\xAA\x9C\xBD\x8E\x1A\xFB\xDFUJ\xB8\xDE\xF5m\xC8Z\xC1\v\\|;\xB6\xE6\xA6\x13o\x91\xE1\xBA\x9An\x8B\x01W\xB9\"\x12+\x9B[\xC7\xCD\xC2O+\f^|\xE1G\x86.%m\xB2x\xF1j\x1B\"\xE41Bu_\x16J\xE4M\x7F\xC5U[\xB9\xD0\xC7\xECu\xE53\xB4)\xD5\x80|\xB7.\xD2\f\xF3\x16\x8B\xE3\x13\x81\x02:5\x88\x0F")
    var unpacked UnpackedThing

    err := binary.Read(input, binary.LittleEndian, &unpacked)
    if err != nil {
        log.Fatal(err)
    }

    log.Printf("[%d, %d, %d, %f, %f]
",
        unpacked.First, unpacked.Second, unpacked.Third, unpacked.Fourth, unpacked.Fifth)
}

We're creating a struct that properly represents the binary format, then using binary.Read() to read our bytes into this struct. One thing to note is that golang's strings don't understand \e, so it must be converted to \x1b.

Output:

*@x~/go/src/stackoverflow.com/binunpack⦕ go run main.go
2018/10/07 17:39:16 [1073741917, 13798400, 0, 2417851639229258349412352.000000, 58518597001656773736660992.000000]
*@x~/go/src/stackoverflow.com/binunpack⦕