* / - 你如何使用+实现XOR？ [英] How do you implement XOR using +-*/?

问题描述

如何XOR操作（两个32位整数）可以只使用基本的算术运算来实现？你有做按位依次通过各2个功率分配后，还是有捷径吗？我不关心执行速度这么多介绍一个最简单，最短code。

How can the XOR operation (on two 32 bit ints) be implemented using only basic arithmetic operations? Do you have to do it bitwise after dividing by each power of 2 in turn, or is there a shortcut? I don't care about execution speed so much as about the simplest, shortest code.

编辑：
这不是功课，而是一个谜上 hacker.org 构成。问题的关键是实现具有非常有限的操作基于堆栈的虚拟机上的XOR（类似于 brainfuck 的语言和肯定的 - 无移位或MOD）。使用VM是困难的部分，当然​​虽然由算法是简短变得更加容易。

This is not homework, but a riddle posed on a hacker.org. The point is to implement XOR on a stack-based virtual machine with very limited operations (similar to the brainfuck language and yes - no shift or mod). Using that VM is the difficult part, though of course made easier by an algorithm that is short and simple.

虽然FryGuy的解决方案是聪明的，我得去与我当初的理想（类似于litb的解决方案），因为比较难以在环境中使用为好。

While FryGuy's solution is clever, I'll have to go with my original ideal (similar to litb's solution) because comparisons are difficult to use as well in that environment.

推荐答案

我很抱歉，我只知道向前伸直之一头：

I'm sorry i only know the straight forward one in head:

uint32_t mod_op(uint32_t a, uint32_t b) {
    uint32_t int_div = a / b;
    return a - (b * int_div);
}

uint32_t xor_op(uint32_t a, uint32_t b) {
    uint32_t n = 1u;
    uint32_t result = 0u;
    while(a != 0 || b != 0) {
        // or just: result += n * mod_op(a - b, 2);
        if(mod_op(a, 2) != mod_op(b, 2)) {
            result += n;
        }
        a /= 2;
        b /= 2;
        n *= 2;
    }
    return result;
}

在注释的备选可用于代替如果避免支化。不过话又说回来了，解决的办法是不完全要么快，这使得它看起来陌生人:)

The alternative in comments can be used instead of the if to avoid branching. But then again, the solution isn't exactly fast either and it makes it look stranger :)

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