为什么在C语言中永远计数一个无符号的int循环? [英] Why does counting down an unsigned int loop forever in C?

问题描述

以下代码在无限循环中运行. 将'i'初始化为值1，然后将其与0进行比较.

The below code runs in an infinte loop. 'i' has been intialised with the value 1 and then compared with 0.

因此，printf()stmt应该执行一次，但可以无限次运行.

So printf() stmt should execute once but it runs infinetly.

unsigned int i = 1;
for (; i >= 0; i--) {
    printf("Hello: %u\n",i);
}

请解释这种行为.

推荐答案

正如其他答案所说，这是因为它是未签名的.我将告诉您一种使用无符号整数 进行操作的优雅方法.

As other answers said, it's because it's unsigned and all. I will tell you an elegant way to do what you want to do with an unsigned integer.

unsigned int i=10;
while(i --> 0) printf("Hello:%u\n", i+1);

此-->有时也称为 运算符.但这实际上只是--和>.如果您更改间距，则会得到

This --> is referred to sometimes as the goes to operator. But it's in fact just -- and >. If you change the spacing, you'll get

while( i-- > 0 )

My2c

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