c++ - 当变量达到最大值时，unsigned int 的递增会导致未定义的行为吗

Question

我的代码中有一个计数器，当它达到 unsigned int 最大值时，我希望我的计数器回到 0。我用一小段代码进行了测试，它可以工作，但我不知道这是否是未定义的行为

#include <stdio.h>
#include <string.h>

main()
{
  unsigned int a = 0;
  a= ~a; // Max value of unsigned int 
  printf("%u \n", a );
  a= a+1; //is it allowed to increment "a" when "a" reach the Max ? 
  printf("%u \n", a ); // display 0

}

Answer 1

a= a+1; //is it allowed to increment "a" when "a" reach the Max ? 

是的，无符号整数永远不会溢出(这是 C 术语)。所以 UINT_MAX + 1 是定义的行为，并且计算为 0。

(C99, 6.2.5p9) "A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type."