英文:
Decoding data from a byte slice to Uint32
问题
package main
import (
"bytes"
"encoding/binary"
"fmt"
)
func main() {
aa := uint(0xFFFFFFFF)
fmt.Println(aa)
byteNewbuf := []byte{0xFF, 0xFF, 0xFF, 0xFF}
buf := bytes.NewBuffer(byteNewbuf)
tt, _ := binary.ReadUvarint(buf)
fmt.Println(tt)
}
英文:
package mainimport ("bytes""encoding/binary""fmt")func main() {aa := uint(0xFFFFFFFF)fmt.Println(aa)byteNewbuf := []byte{0xFF, 0xFF, 0xFF, 0xFF}buf := bytes.NewBuffer(byteNewbuf)tt, _ := binary.ReadUvarint(buf)fmt.Println(tt)}
Need to convert 4 bytes array to uint32 but why the results are not same ?
go verion : beta 1.1
答案1
得分: 13
你可以使用encoding/binary包中的ByteOrder对象之一来完成这个操作。例如：
package mainimport ("encoding/binary""fmt")func main() {aa := uint(0x7FFFFFFF)fmt.Println(aa)slice := []byte{0xFF, 0xFF, 0xFF, 0x7F}tt := binary.LittleEndian.Uint32(slice)fmt.Println(tt)}
如果你的数据是大端格式的，你可以使用相同的方法在binary.BigEndian上进行操作。
英文:
You can do this with one of the ByteOrder objects from the encoding/binary package. For instance:
package mainimport ("encoding/binary""fmt")func main() {aa := uint(0x7FFFFFFF)fmt.Println(aa)slice := []byte{0xFF, 0xFF, 0xFF, 0x7F}tt := binary.LittleEndian.Uint32(slice)fmt.Println(tt)}
If your data is in big endian format, you can instead use the same methods on binary.BigEndian.
答案2
得分: 5
tt := uint32(buf[0])<<24 | uint32(buf1)<<16 | uint32(buf2) <<8 |
uint32(buf3)
for BE or
tt := uint32(buf[0]) | uint32(buf1)<<8 | uint32(buf2) <<16 |
uint32(buf3) <<24
for LE.
varint是一种不同类型的编码（32位数字在编码形式中可以有多达5个字节，64位数字最多可以有10个字节）。
-
不需要为[]byte创建缓冲区。直接在字节切片上使用Varint或Uvarint。
-
您正在丢弃函数返回的错误。第二个结果指示读取了多少字节或是否存在问题。将0xff、0xff、0xff、0xff解码为uvarint时确实存在问题。
英文:
tt := uint32(buf[0])<<24 | uint32(buf[1])<<16 | uint32(buf[2]) <<8 |uint32(buf[3])
for BE or
tt := uint32(buf[0]) | uint32(buf[1])<<8 | uint32(buf[2]) <<16 |uint32(buf[3]) <<24
for LE.
[u]varint is a different kind of encoding (32 bit numbers can have as much as 5 bytes in the encoded form, 64 bit numbers up to 10).
-
No need to create a buffer for []byte. Use Varint or Uvarint directly on the byte slice instead.
-
You're throwing away the error returned by the function. The second result indicates how many bytes were read or if there was a problem. There is a problem while decoding 0xff, 0xff, 0xff, 0xff as an uvarint.
答案3
得分: 4
这里是如何使用encoding/binary包来实现你想要的功能。请注意，你不应该使用任何var函数，因为它们会进行可变长度编码。
package mainimport ("bytes""encoding/binary""fmt""log")func main() {aa := uint(0xFFFFFF0F)fmt.Println(aa)tt := uint32(0)byteNewbuf := []byte{0x0F, 0xFF, 0xFF, 0xFF}buf := bytes.NewBuffer(byteNewbuf)err := binary.Read(buf, binary.LittleEndian, &tt)if err != nil {log.Fatalf("Decode failed: %s", err)}fmt.Println(tt)}
结果是
42949670554294967055
英文:
Here is how to use the encoding/binary package to do what you want. Note that you don't want to use any of the var functions as those do variable length encoding.
package mainimport ("bytes""encoding/binary""fmt""log")func main() {aa := uint(0xFFFFFF0F)fmt.Println(aa)tt := uint32(0)byteNewbuf := []byte{0x0F, 0xFF, 0xFF, 0xFF}buf := bytes.NewBuffer(byteNewbuf)err := binary.Read(buf, binary.LittleEndian, &tt)if err != nil {log.Fatalf("Decode failed: %s", err)}fmt.Println(tt)}
Result is
42949670554294967055
答案4
得分: -1
> 数字类型
>
> byte 是 uint8 的别名
由于 byte 是 uint8 的别名，你的问题，“需要将 4 字节数组转换为 uint32”，已经有了答案：
如何在 Go 中将 [4]uint8 转换为 uint32？
> 二进制包
>
> [Uvarints 和] Varints 是一种使用一个或多个字节对整数进行编码的方法；绝对值较小的数字占用较少的字节。有关规范，请参见
> http://code.google.com/apis/protocolbuffers/docs/encoding.html。
由于 Uvarint 是一种特殊的整数表示和存储形式，你应该只在使用 Uvarint 函数写入的值上使用 ReadUvarint 函数。
例如，
package mainimport ("bytes""encoding/binary""fmt")func main() {buf := make([]byte, 10)x := uint64(0xFFFFFFFF)fmt.Printf("%2d %2d %v\n", x, len(buf), buf)n := binary.PutUvarint(buf, x)buf = buf[:n]fmt.Printf("%2d %2d %v\n", x, len(buf), buf)y, err := binary.ReadUvarint(bytes.NewBuffer(buf))if err != nil {fmt.Println(err)return}fmt.Printf("%2d %2d %v\n", y, len(buf), buf)}
输出：
4294967295 10 [0 0 0 0 0 0 0 0 0 0]4294967295 5 [255 255 255 255 15]4294967295 5 [255 255 255 255 15]
英文:
> Numeric types
>
> byte alias for uint8
Since byte is an alias for uint8, your question, "Need to convert 4 bytes array to uint32", has already been answered:
How to convert [4]uint8 into uint32 in Go?
> Package binary
>
> [Uvarints and] Varints are a method of encoding integers using one
> or more bytes; numbers with smaller absolute value take a smaller
> number of bytes. For a specification, see
> http://code.google.com/apis/protocolbuffers/docs/encoding.html.
Since Uvarints are a peculiar form of integer representation and storage, you should only use the ReadUvarint function on values that have been written with the Uvarint function.
For example,
package mainimport ("bytes""encoding/binary""fmt")func main() {buf := make([]byte, 10)x := uint64(0xFFFFFFFF)fmt.Printf("%2d %2d %v\n", x, len(buf), buf)n := binary.PutUvarint(buf, x)buf = buf[:n]fmt.Printf("%2d %2d %v\n", x, len(buf), buf)y, err := binary.ReadUvarint(bytes.NewBuffer(buf))if err != nil {fmt.Println(err)return}fmt.Printf("%2d %2d %v\n", y, len(buf), buf)}
Output:
4294967295 10 [0 0 0 0 0 0 0 0 0 0]4294967295 5 [255 255 255 255 15]4294967295 5 [255 255 255 255 15]
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