使用按位运算符的条件语句 [英] Conditional Statement using Bitwise operators

问题描述

因此，我看到已经提出了这个问题，但是答案有些模糊且无济于事.好的，我只需要使用& ^〜！+ | >><<"

So I see that this question has already been asked, however the answers were a little vague and unhelpful. Okay, I need to implement a c expression using only "& ^ ~ ! + | >> <<"

表达式需要类似于:a? b:c

The expression needs to resemble: a ? b : c

因此，据我所知，表达式需要看起来像这样:

So, from what I've been able to tell, the expression needs to look something like:

return (a & b) | (~a & c)

这在a = 0时起作用，因为将其与b取和将得到零，然后or表达式将返回右侧，(~a & c)起作用，因为〜0给出全为1，而将所有c都与和c则返回c

This works when a = 0, because anding it with b will give zero, and then the or expression will return the right side, (~a & c) which works because ~0 gives all ones, and anding c with all ones returns c.

但是，当>> 0时，此功能将无效.有人可以尝试解释它的原因或解决方法吗?

However, this doesn't work when a > 0. Can someone try to explain why this is, or how to fix it?

推荐答案

我将使用!!a将a转换为布尔值，以获取0或1.x = !!a.

I would convert a to a boolean using !!a, to get 0 or 1. x = !!a.

然后我将其取二.由于没有可用的一元减号，因此可以使用2的补数求反的定义:反转位，然后加一个:y = ~x + 1.这样将清除所有位或设置所有位.

Then I'd negate that in two's complement. Since you don't have unary minus available, you use the definition of 2's complement negation: invert the bits, then add one: y = ~x + 1. That will give either all bits clear, or all bits set.

然后我将直接使用一个变量y & b与另一个变量~y & c取反的and.这将为其中一个表达式赋予0，为另一个表达式赋予原始变量.当我们将它们全部or在一起时，零将无效，因此我们将获得原始变量，不变.

Then I'd and that directly with one variable y & b, its inverse with the other: ~y & c. That will give a 0 for one of the expressions, and the original variable for the other. When we or those together, the zero will have no effect, so we'll get the original variable, unchanged.

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