确定32位int的符号 [英] Determine the sign of a 32 bit int

问题描述

仅使用:

！ 〜& ^ | +<< >>

! ~ & ^ | + << >>

没有麻烦

我需要确定32位整数的正负号，如果为正，则需要返回1，如果为0，则需要返回0，如果为负，则需要返回-1.

I need to determine the sign of a 32 bit integer and I need to return 1 if positive, 0 if 0 and -1 if negative.

有什么想法吗?我首先考虑过移31位，然后看那个符号，但是那显然不起作用，现在我有点卡住了.

Any ideas? I first thought about shifting over 31 bits and then looking at that sign but that obviously wont work and now I am kind of stuck.

推荐答案

尝试一下:

(x >> 31) | (((0 - x) >> 31) & 1)

如何?

(x >> 31) | (((~x + 1) >> 31) & 1)

针对评论中提出的问题(或更确切地说是挑剔的问题)...

In response to issues (or rather nit-picking) raised in the comments...

这些解决方案有效的假设:

Assumptions for these solutions to be valid:

1. x是32位有符号整数类型.
2. 在此系统上，带符号的32位整数是二进制补码. (右移是算术运算)
3. 算术溢出的环绕式.
4. 对于第一个解决方案，文字0与x的类型相同.

1. x is of type 32-bit signed integer.
2. On this system, signed 32-bit integers are two's complement. (right-shift is arithmetic)
3. Wrap-around on arithmetic overflow.
4. For the first solution, the literal 0 is the same type as x.

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