在Go中将uint16强制转换为int16的正确方法 [英] Proper way for casting uint16 to int16 in Go

问题描述

按位操作和Go新手:D我正在用Go从传感器读取一些数据，并以2个字节的形式获取它-假设 0xFFFE .也很容易将其转换为uint16，因为在Go中我们可以执行 uint16(0xFFFE)，但是我需要将其转换为整数，因为传感器实际上返回的值范围是-32768到32767.现在我想也许Go会很不错，如果我做 int16(0xFFFE)，它将明白我想要什么?" ，但没有.我最终使用了以下解决方案(我从网络上翻译了一些python代码):

Bitwise manipulation and Go newbie here :D I am reading some data from sensor with Go and I get it as 2 bytes - let's say 0xFFFE. It is easy too cast it to uint16 since in Go we can just do uint16(0xFFFE) but what I need is to convert it to integer, because the sensor returns in fact values in range from -32768 to 32767. Now I thought "Maybe Go will be this nice and if I do int16(0xFFFE) it will understand what I want?", but no. I ended up using following solution (I translated some python code from web):

x := 0xFFFE

if (x & (1 << 15)) != 0 {
    x = x - (1<<16)
}

这似乎可行，但是A)我不确定为什么，B)我认为应该是将uint16转换为int16的简单解决方案，这看起来有点难看.谁能帮我一下，阐明为什么这是唯一的方法吗?还是还有其他可能的方法?

It seems to work, but A) I am not entirely sure why, and B) It looks a bit ugly to what I imagined should be a trivial solution for casting uint16 to int16. Could anyone give me a hand and clarify why this is only way to do this? Or is there any other possible way?

推荐答案

但是您想要的是，"Go很好":

But what you want works, "Go is nice":

ui := uint16(0xFFFE)
fmt.Println(ui)
i := int16(ui)
fmt.Println(i)

输出(在游乐场上尝试):

65534
-2

int16(0xFFFE)不起作用，因为 0xfffe 是一个无类型的整数常量，不能用 int16 类型的值表示，这就是编译器抱怨的原因.但是您当然可以将任何 uint16 非恒定值转换为 int16 .

int16(0xFFFE) doesn't work because 0xfffe is an untyped integer constant which cannot be represented by a value of type int16, that's why the the compiler complains. But you can certainly convert any uint16 non-constant value to int16.

查看可能的重复项:

Golang:故意的int溢出

可以使用编译器的常量表达式和其他表达式的评估方式不同

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