INT_MAX+1 = INT_MIN 是有符号整数吗? [英] Is INT_MAX+1 = INT_MIN in signed integer?

问题描述

for (i = 0; i <= N; ++i) { ... }

如果 N 是 INT_MAX，则此特定语句将导致无限循环.知道无符号溢出正在包装溢出，假设 i 和 N 为无符号，编译器可以假设循环将精确迭代 N+1 次如果 i 在溢出时未定义.这里要注意的是:如果我将循环设为，

This particular statement will cause an infinite loop if N is INT_MAX. Having known that Unsigned Overflows are wrapping overflows, assuming i and N to unsigned, compiler can assume that the loop will iterate exactly N+1 times if i is undefined on overflow. The thing to note here is: if I make the loops as,

for (i = 0; i < N; ++i) { ... }

这仍然是未定义的行为吗?

Will this still be undefined behav?

在有符号整数的情况下，为什么 INT_MAX + 1 不一定等于 INT_MIN?

Why INT_MAX + 1 is not surely equal to INT_MIN in case of signed integers?

推荐答案

INT_MAX + 1

此操作调用未定义的行为.有符号整数溢出是 C 中未定义的行为.

this operation invokes undefined behavior. Signed integer overflow is undefined behavior in C.

它可能导致 INT_MIN 或者实现可以认为这个表达式是正的，否则程序会崩溃.不要让便携式程序计算这个表达式.

It can result to INT_MIN or the implementation can consider this expression to be positive or the program can crash. Do not let a portable program compute this expression.

这篇关于INT_MAX+1 = INT_MIN 是有符号整数吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！