有条件的使用按位运算符 [英] Conditional Using Bitwise Operators

问题描述

如何使用按位运算符表示条件运算符?

How is the conditional operator represented using bitwise operators?

这是一个作业问题，在这里我必须仅使用按位运算来实现条件运算符.如果允许使用if语句会很简单，但是必须严格按位进行运算.

It is a homework question where I have to implement the conditional operator using only bitwise operations. It would be simple if if statements were allowed, however it has to be strictly bitwise operators.

只能使用运算符!，~，&，^，|，+，>>和<<.不能使用if语句或循环.

Only the operators !, ~, &, ^, |, +, >>, and << can be used. No if statements or loops can be used.

该函数采用三个整数，并且与普通的条件运算符一样工作.第一个参数的值为零或非零.如果第一个参数为零，则返回第二个参数.如果第一个参数不为零，则返回第三个参数.

The function takes three ints and works just like the normal conditional operator. The first argument is evaluated as either zero or non-zero. If the first argument is zero then the second argument is returned. If the first argument is non-zero then the third argument is returned.

我希望对此有一个简单的算法.关于从哪里开始的任何想法都会有很大的帮助.

I was hoping there would be a simple algorithm for this. Any ideas on where to start would be a great help.

推荐答案

是否允许将移位作为按位运算符?可以使用算术运算符吗?

Are shifts allowed as bitwise operators? Are arithmetic operators allowed?

您的修改尚不完全清楚，但是我认为您需要实现等效的

Your edit is not entirely clear, but I assume that you need to implement an equivalent of

a ? b : c

其中a，b和c是整数.反过来等同于

where a, b and c are integers. This is in turn equivalent to

a != 0 ? b : c

一种实现方法是找到一种仅使用按位运算符将a的非零值转换为全1的位模式的方法.如果我们弄清楚该怎么做，其余的将很容易.现在，我不会立即记住任何能做到这一点的巧妙技巧(我相信它们确实存在)，而且我不确定是否允许哪些运算符，所以现在我将只使用类似

One way to achieve that is to find a way to turn non-zero value of a into an all-ones bit pattern using only bitwise operators. If we figure out how to do that, the rest would be easy. Now, I don't immediately remember any ingenious tricks that would do that (they do exist I believe), and I don't know for sure which operators are allowed and which are not, so for now I will just use something like

a |= a >> 1; a |= a >> 2; a |= a >> 4; a |= a >> 8; a |= a >> 16;
a |= a << 1; a |= a << 2; a |= a << 4; a |= a << 8; a |= a << 16;

对于32位整数类型，如果(且仅当)原始a中至少设置了一位，以上内容应导致a的所有位均设置为1.(假设我们正在使用无符号整数，以避免与有符号值移位相关的问题.再次，我敢肯定，必须有一种更聪明的方法来做到这一点.例如:a = !a - 1，但是我不知道是否允许!和-.

For a 32-bit integer type, if (and only if) there was at least one bit set in the original a, the above should result in all bits of a set to 1. (Let's assume we are working with unsigned integers, to avoid the issues associated with shifting of signed values). Again, there must be a more clever way to do that, I'm sure. For example: a = !a - 1, but I don't know if ! and - are allowed.

完成此操作后，原始条件运算符将等效于

Once we've done that, the original conditional operator becomes equivalent to

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

完成.

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