c - c中的翻转位

Question

在这段代码中，我应该要求用户翻转 32 位中的 1。当我翻转第 32 位(索引 31)时，即使我使用无符号整数，我也会收到 -2147483648 。如何修复翻盖才能正确伸出？

int main(int argc, char *argv[]) {
   int32_t num = 0;
   int bitC = 0;
   int number;
   char cont;
   do {
      printf("Enter a number between 1 and 1000\n");
      do {
         scanf("   %d", &num);
         if(num < 0 || num > 1000) {
            printf("Error: Number MUST be between 1 and 1000.\n");
            //printf("Please re-enter number.\n");
         }
      } while(num < 0 || num > 1000);
      printf("Choose a bit, between 0 and 31, to flip\n");
      do {
         scanf("   %d", &bitC);
         if(bitC < 0 || bitC > 31) {
            printf("Error: bit MUST be between 0 and 31\n");
            //printf("Please re-enter bit to change.\n");
         }
      } while(bitC <0 || bitC > 31);
      printf("Number before bit flip was %d\n", num);
      //num ^= (-bitC ^ num)&(1 << bitC);
      num ^= (1UL << bitC);
      printf("New number is: %d\n", num);
      printf("Would you like to shift another bit?\n");
      printf("Enter y to continue or n to quit\n");
      scanf("   %s", &cont);
   } while(cont == 'y');
   return 0;
}

Answer 1

您现在正在程序中使用有符号整数。但真正的问题是，当你 printf 时，你使用的是 %d，这会导致 printf 将参数求值(记住 printf 不知道它到底是什么，它只是堆栈上的一个值)作为有符号 int，即使传递给 printf 的值是无符号的。使用%u。