我如何检查是否一个整数是奇数还是偶数？ [英] How do I check if an integer is even or odd?

问题描述

我如何检查，如果给定的数字是奇数还是偶数用C？

How can I check if a given number is even or odd in C?

推荐答案

使用模（％）运算符来检查是否有一个余数除以2时：

Use the modulo (%) operator to check if there's a remainder when dividing by 2:

if (x % 2) { /* x is odd */ }

一些人批评上述声明我的答案用x＆放大器; 1是更快或更高效。我不相信这是事实。

A few people have criticized my answer above stating that using x & 1 is "faster" or "more efficient". I do not believe this to be the case.

出于好奇，我创建了两个平凡的测试用例方案：

Out of curiosity, I created two trivial test case programs:

/* modulo.c */
#include <stdio.h>

int main(void)
{
    int x;
    for (x = 0; x < 10; x++)
        if (x % 2)
            printf("%d is odd\n", x);
    return 0;
}

/* and.c */
#include <stdio.h>

int main(void)
{
    int x;
    for (x = 0; x < 10; x++)
        if (x & 1)
            printf("%d is odd\n", x);
    return 0;
}

然后我用gcc 4.1.3在我的机器之一，5个不同的时间编译的这些：

I then compiled these with gcc 4.1.3 on one of my machines 5 different times:

• 由于没有优化的标志。


• 用-O


• 使用-Os


• 随着-O2


• 使用-O3

我检查的每个编译的汇编输出（使用gcc -S），发现在每一种情况下，输出为and.c和modulo.c是相同的（它们都使用了和L $ 1，％eax中的指令）。我怀疑这是一个新的特点，让我怀疑它的历史可以追溯到古老的版本。我也怀疑任何现代（在过去的20年间）的非神秘的编译器，商业或开源，缺乏这种优化。我将测试在其他的编译器，但我没有任何可用的时刻。

I examined the assembly output of each compile (using gcc -S) and found that in each case, the output for and.c and modulo.c were identical (they both used the andl $1, %eax instruction). I doubt this is a "new" feature, and I suspect it dates back to ancient versions. I also doubt any modern (made in the past 20 years) non-arcane compiler, commercial or open source, lacks such optimization. I would test on other compilers, but I don't have any available at the moment.

如果任何人会关心测试其他的编译器和/或平台的目标，并得到了不同的结果，我会很有兴趣知道。

If anyone else would care to test other compilers and/or platform targets, and gets a different result, I'd be very interested to know.

最后，模版本的保证按标准工作整数是正数，负数或零，不管有符号整数实施的再presentation的。按位和版本是没有的。是的，我知道补多少有些无处不在，所以这不是一个真正的问题。

Finally, the modulo version is guaranteed by the standard to work whether the integer is positive, negative or zero, regardless of the implementation's representation of signed integers. The bitwise-and version is not. Yes, I realise two's complement is somewhat ubiquitous, so this is not really an issue.

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