对于循环条件使用无符号整数时，停在0？ [英] For loop condition to stop at 0 when using unsigned integers?

问题描述

我有了从n转到0（（含））的循环。我的 I 变量的类型为size_t 通常是无符号的。我目前使用以下code：

I have a loop that has to go from N to 0 (inclusively). My i variable is of type size_t which is usually unsigned. I am currently using the following code:

for (size_t i = N; i != (size_t) -1; --i) {
    ...
}

这是否正确？有没有更好的方式来处理的条件？

Is that correct? Is there a better way to handle the condition?

谢谢，

文森特。

推荐答案

是的，这是正确的，这是一个非常普遍的做法。我不会考虑改变它。

Yes, it's correct and it is a very common approach. I wouldn't consider changing it.

这是无符号整型运算是保证使用模 2 ^ N 算术（其中 N 是多少在类型值位）和溢出行为明确界定。结果被转换成 0 的范围为 2 ^ N - 1 通过增加或减少<$ C $的倍数C> 2 ^ N （即模 2 ^ N 算术）。

Arithmetic on unsigned integer types is guaranteed to use modulo 2^N arithmetic (where N is the number of value bits in the type) and behaviour on overflow is well defined. The result is converted into the range 0 to 2^N - 1 by adding or subtracting multiples of 2^N (i.e. modulo 2^N arithmetic).

1 转换为无符号整型（其中为size_t 就是其中之一）转换为 2 ^ N - 1 。 - 还采用模 2 ^ N 算术无符号类型，因此无符号类型与值 0 将递减到 2 ^ N - 1 。你的循环终止条件是正确的。

-1 converted to an unsigned integer type (of which size_t is one) converts to 2^N - 1. -- also uses modulo 2^N arithmetic for unsigned types so an unsigned type with value 0 will be decremented to 2^N - 1. Your loop termination condition is correct.

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