英文:
How to write LEB128 in Go
问题
如何在Go中将整数写入LEB128格式？我正在尝试将int32编码为Minecraft VarInt，到目前为止，我已经尝试将维基百科上的示例导入到Go中。但是在测试时，我得到了错误的结果，维基百科上说-1应该等于[255 255 255 255 15]，但我得到的是[255 255 255 255 255]。我在这里做错了什么？
func WriteVarInt2(v int32) []byte{var out []bytec := 0for{currentByte := byte(v & 0b01111111)v >>= 7if v != 0 {currentByte |= 0b10000000}out = append(out, currentByte)c++if c >= 5 || v == 0{return out}}}
英文:
How do you write an integer to LEB128 format in Go? I'm trying to encode an int32 to a Minecraft VarInt, so far I've tried importing the example on the wiki to Go. I get the wrong results when testing though, wiki says -1 should equal [255 255 255 255 15], but I get [255 255 255 255 255] instead. What I'm I doing wrong here?
func WriteVarInt2(v int32) []byte{var out []bytec := 0for{currentByte := byte(v & 0b01111111)v >>= 7if v != 0 {currentByte |= 0b10000000}out = append(out, currentByte)c++if c >= 5 || v == 0{return out}}}
答案1
得分: 6
问题出在移位操作上。
>> 是算术右移，>>> 是逻辑右移。它们的区别在于 >> 会带入符号位（在左边），而 >>> 会带入零（无论符号位是什么）。
LEB128 变长整数的算法使用的是逻辑右移，而 Go 的 >> 是算术右移。
在 Go 中没有明确的逻辑右移操作，但是如果将数字视为无符号数，你将得到逻辑右移的效果：
func WriteVarInt2(v_ int32) []byte {v := uint32(v_)// 函数的其余部分保持不变// ...}
进行测试：
fmt.Println(WriteVarInt2(-1))
输出结果如预期（在 Go Playground 上尝试）：
[255 255 255 255 15]
英文:
The problem is with the shifting operation.
>> is arithmetic shift right, >>> is logical shift right. The difference is that >> brings in the sign bit (on the left), while >>> brings in zeros (whatever the sign bit was).
The algorithm of LEB128's Varint uses logical shift, and Go's >> is arithmetic shift.
There is no distinct logical shift in Go, but if you treat the number as unsigned, you'll get exactly that:
func WriteVarInt2(v_ int32) []byte {v := uint32(v_)// rest of your function unchanged// ...}
Testing it:
fmt.Println(WriteVarInt2(-1))
Output is as expected (try it on the Go Playground):
[255 255 255 255 15]
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