同为size_t比较INT [英] comparing int with size_t

问题描述

如果我有一个int和一个为size_t变量，我可以比较它们，如：

If i have a int and a size_t variable,can i compare them like:

int i=1;
size_t y=2;
if(i==y)
do something..

或我要类型转换其中之一吗？

or i have to type-cast one of them?

推荐答案

它的安全提供的 INT 是零或正值。如果是负的，而为size_t 等于或更高级别的比 INT ，那么 INT 将被转换为为size_t 等其负面价值将反而成为正值。然后，这个新的积极的价值是相对于为size_t 值，它可以（在一个惊人的巧合不太可能）给假阳性。要真正安全的（也许是缩手缩脚）检查 INT 非负第一：

It's safe provided the int is zero or positive. If it's negative, and size_t is of equal or higher rank than int, then the int will be converted to size_t and so its negative value will instead become a positive value. This new positive value is then compared to the size_t value, which may (in a staggeringly unlikely coincidence) give a false positive. To be truly safe (and perhaps overcautious) check that the int is nonnegative first:

/* given int i; size_t s; */
if (i>=0 && i == s)

和共进晚餐preSS编译器警告：

and to suppress compiler warnings:

if (i>=0 && (size_t)i == s)

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