dongpangfu6322 2018-10-07 20:51
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Golang中的字符串解码

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]

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1条回答 默认 最新

  • duanqun7761 2018-10-08 00:41
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    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⦕
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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