int型变量创建上溢下溢并在C中输出值

Question

我必须编写一个程序来为int类型的变量创建上溢和下溢，并在发生上溢或下溢时以十进制和十六进制输出变量的值。

这是我的代码:

#include <stdio.h>
#include <limits.h>

int main() {


   int delta = INT_MAX;
   printf("\n delta = %d (%4x)", delta, delta);
   delta = delta + 1;
   printf("\n delta +1 = %d (%4x)", delta, delta);
   delta = delta +2;
   printf("\n delta +2 = %d (%4x)", delta, delta);

   printf("\n");

   int delta_min = INT_MIN;
   printf("\n delta_min = %d (%4x)", delta_min, delta_min);
   delta_min = delta_min - 1;
   printf("\n delta_min -1 = %d (%4x)", delta_min, delta_min);
   delta_min = delta_min -2;
   printf("\n delta_min -2 = %d (%4x) \n\n", delta_min, delta_min);

return 0;
}

答案应该是这样的

delta         = 2147483647   (7fffffff)
delta +1      = -2147483648  (80000000)
delta + 2     = -2147483647  (80000001)

delta_min     = -2147483648  (80000000)
delta_min -1  = 2147483647   (7fffffff)
delta_min -2  = 2147483646   (7ffffffe)

但是我的回答是这样的

delta         = 2147483647   (7fffffff)
delta +1      = -2147483648  (80000000)
delta + 2     = -2147483646  (80000002)    <<

delta_min     = -2147483648  (80000000)
delta_min -1  = 2147483647   (7fffffff)
delta_min -2  = 2147483645   (7ffffffd)    <<

我不知道我在这里做错了什么。我不是要你解决我的家庭作业。只要一个提示就会非常有帮助!

感谢您抽空阅读~

Answer 1

您似乎不确定为什么要加 3 而不是加 2。

delta + 2     = -2147483646  (80000002)    <<

这是因为您重新分配了 delta 变量，而不是将结果存储在单独的变量中。

delta = delta + 1 // delta is reassigned to -2147483648
// delta + 1     = -2147483648  (80000000)
delta = delta + 2 // Here is the problem. It should be delta + 1.
// delta + 2     = -2147483646  (80000002)    <<

相反，您应该每次只加 1。