在Go中将uint16强制转换为int16的正确方法

Question

Bitwise manipulation and Go newbie here :DI 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)
}

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?

Answer 1

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

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

Output (try it on the Go Playground ):

65534
-2

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 .

See possible duplicates:

Golang: on-purpose int overflow

Does go compiler's evaluation differ for constant expression and other expression