c - 如何解决 c 中的 int 溢出问题

Question

在 C 语言中，是否可以在不导致整数溢出的情况下执行以下操作？ 我的答案需要是一个整数，我稍后会在程序中使用它。代码打印-337。 正确答案应该是 2014 年。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main() {
   int i;
   i = (10000 * 735625 + 14780)/3652425;
   printf("%d",i);
   return 0;
}

Answer 1

基于usual arithmetic conversions ，当 10000 * 735625 + 14780(超出 int 的范围)尝试保存到 int 时，它将溢出。

Integral promotions are performed on the operands as follows:

• If either operand is of type unsigned long, the other operand is converted to type unsigned long.
• If preceding condition not met, and if either operand is of type long and the other of type unsigned int, both operands are converted to type unsigned long.
• If the preceding two conditions are not met, and if either operand is of type long, the other operand is converted to type long.
• If the preceding three conditions are not met, and if either operand is of type unsigned int, the other operand is converted to type unsigned int.
• If none of the preceding conditions are met, both operands are converted to type int.

要计算，您需要使用long long(范围更大)。

正如 @JoachimIsaksson 所建议的，一种简单的方法是将 LL 放在 10000 之后，以 long long 的精度进行计算:

i = ( 10000LL * 735625 + 14780)/3652425;

现场观看:http://ideone.com/pA2Pvm