英文:
Converting between a 3-byte slice and signed integer type
问题
我有一个由3个字节组成的切片（按照小端序排列），表示一个有符号整数，我想将其转换为任何整数类型，最好是int32，然后再转回原来的形式。
b := []byte{0x01, 0x00, 0x80}
我尝试使用big.Int来实现，但似乎它的SetBytes()和Bytes()方法只适用于无符号整数类型，而不管big.Int实例也可以存储有符号整数类型。
以下是我尝试实现的Python等效代码：
b = b"\x01\x00\x80"
i = int.from_bytes(b, "little", signed=True)
b_again = int.to_bytes(i, "little", signed=True)
编辑：这个问题不是https://stackoverflow.com/questions/34701187/go-byte-to-little-big-endian-signed-integer-or-float的重复。我尝试了那里的答案，但结果并不如预期。请参考我的第一个评论。
英文:
I have a slice consisting of 3 bytes (ordered in LE) that is representing a signed integer, and I want to convert it to any of the integer types, preferably int32, and then back to itself.
b := []byte{0x01, 0x00, 0x80}
I tried to do that using big.Int, but it seemed like its SetBytes() and Bytes() methods only work for unsigned integer types, regardless of that a big.Int instance can also store signed integer types.
Here is a Python equivalent of what I’m trying to do:
b = b"\x01\x00\x80"
i = int.from_bytes(b, "little", signed=True)
b_again = int.to_bytes("little", signed=True)
Edit: This question is not a duplicate of https://stackoverflow.com/questions/34701187/go-byte-to-little-big-endian-signed-integer-or-float. I have tried applying the answers there, but their results weren’t expected. See my first comment.
答案1
得分: 3
我有一个由3个字节组成的切片（按照小端序排列），表示一个有符号整数，我想将其转换为任何整数类型，最好是int32，然后再转回原来的类型。
package main
import "fmt"
func ByteLE3ToInt24(b []byte) int32 {
i := uint32(b[0])<<8 | uint32(b[1])<<16 | uint32(b[2])<<24
return int32(i) >> 8
}
func Int24ToByteLE3(i int32) []byte {
return []byte{byte(i), byte(i >> 8), byte(i >> 16)}
}
func main() {
b := []byte{0x01, 0x00, 0x80} // 负数
fmt.Println(b)
fmt.Println(ByteLE3ToInt24(b))
fmt.Println(Int24ToByteLE3(ByteLE3ToInt24(b)))
fmt.Println()
b = []byte{0x01, 0x00, 0x00} // 正数
fmt.Println(b)
fmt.Println(ByteLE3ToInt24(b))
fmt.Println(Int24ToByteLE3(ByteLE3ToInt24(b)))
}
输出结果：
[1 0 128]
-8388607
[1 0 128]
[1 0 0]
1
[1 0 0]
你可以在这里查看代码和运行结果：https://go.dev/play/p/tI8E2kSXopZ
英文:
> I have a slice consisting of 3 bytes (ordered in LE) that is representing a signed integer, and I want to convert it to any of the integer types, preferably int32, and then back to itself.
package main
import "fmt"
func ByteLE3ToInt24(b []byte) int32 {
i := uint32(b[0])<<8 | uint32(b[1])<<16 | uint32(b[2])<<24
return int32(i) >> 8
}
func Int24ToByteLE3(i int32) []byte {
return []byte{byte(i), byte(i >> 8), byte(i >> 16)}
}
func main() {
b := []byte{0x01, 0x00, 0x80} // Negative
fmt.Println(b)
fmt.Println(ByteLE3ToInt24(b))
fmt.Println(Int24ToByteLE3(ByteLE3ToInt24(b)))
fmt.Println()
b = []byte{0x01, 0x00, 0x00} // Positive
fmt.Println(b)
fmt.Println(ByteLE3ToInt24(b))
fmt.Println(Int24ToByteLE3(ByteLE3ToInt24(b)))
}
https://go.dev/play/p/tI8E2kSXopZ
[1 0 128]
-8388607
[1 0 128]
[1 0 0]
1
[1 0 0]
答案2
得分: -1
Little-Endian（小端序）意味着数字0x01020304被编码为字节序列[]byte{4, 3, 2, 1}。
你的示例数字有三个字节，所以我假设你的意思是0x00010080（左侧填充一个零字节）。
有一个encoding/binary.ByteOrder类型，它有两个实例LittleEndian和BigEndian，提供了转换为字节和从字节转换的函数：
package main
import (
"encoding/binary"
"fmt"
)
func main() {
var b [4]byte
binary.LittleEndian.PutUint32(b[:], 0x00010080)
i32 := int32(binary.LittleEndian.Uint32(b[:]))
fmt.Println(b) // 输出 [128 0 1 0]。
fmt.Println(i32) // 输出 65664（等于0x00010080）。
}
英文:
Little-Endian means that the number 0x01020304 is encoded as the byte squence []byte{4, 3, 2, 1}.
Your example number has three bytes, so I assume that you mean 0x00010080 (left-filled with a zero-byte).
There is a type encoding/binary.ByteOrder with the two instances LittleEndian and BigEndian which provide functions to convert to and from bytes:
package main
import (
"encoding/binary"
"fmt"
)
func main() {
var b [4]byte
binary.LittleEndian.PutUint32(b[:], 0x00010080)
i32 := int32(binary.LittleEndian.Uint32(b[:]))
fmt.Println(b) // Prints [128 0 1 0].
fmt.Println(i32) // Prints 65664 (which is 0x00010080).
}
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