doulan9287
2019-02-21 19:01
浏览 893
已采纳

在Golang中将二进制值作为字符串转换为uint32

Hello i am trying to convert 00000000000000000000000000001011 to uint32 in golang using

var v = "00000000000000000000000000001011"
fmt.Printf("%T
", v)
c := []byte(v)
u := binary.LittleEndian.Uint32(c)

However it is not working.

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您好,我正在尝试使用

 <将go000000中的00000000000000000000000000000000001011转换为uint32 代码> var v =“ 00000000000000000000000000001011” 
fmt.Printf(“%T 
”,v)
c:= [] byte(v)
u:= binary.LittleEndian.Uint32(c)
    
 
 

但是它不起作用。

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2条回答 默认 最新

  • dou47732 2019-02-21 19:16
    已采纳

    You can't use encoding/binary for this, as that is to serialize and deserialize the (memory) bytes of different values (e.g. numbers). What you have is the base 2 string representation of the number.

    To get its integer value you have to parse it. For that, use strconv.ParseUint():

    s := "00000000000000000000000000001011"
    u, err := strconv.ParseUint(s, 2, 32)
    if err != nil {
        panic(err)
    }
    fmt.Println(u)
    

    This outputs (try it on the Go Playground):

    11
    

    Note that strconv.ParseUint() returns a value of type uint64, so if you need uint32, you have to manually convert it, e.g.:

    u32 := uint32(u)
    

    There are more options for parsing numbers from strings, for an overview, check Convert string to integer type in Go?

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  • dtsps00544 2019-02-21 19:18

    For example,

    package main
    
    import (
        "fmt"
        "strconv"
    )
    
    func main() {
        s := "00000000000000000000000000001011"
        fmt.Println(s)
        u64, err := strconv.ParseUint(s, 2, 32)
        u32 := uint32(u64)
        if err == nil {
            fmt.Println(u32)
        }
    }
    

    Playground: https://play.golang.org/p/yiicgWsb7B_M

    Output:

    00000000000000000000000000001011
    11
    
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