Question

尝试在我的代码中断中编译以下行：

printf("%llu\n", 0x12345678 * 0x12345678);

我明白了：

program.c:45:34: warning: integer overflow in expression [-Woverflow]
     printf("%llu\n", (0x12345678 * 0x12345678));

我该如何解决这个问题？

Answer 1

[按照@Lundin]评论

接受更正后

在您的计算机上，0x12345678比unsigned long long更窄 - 当然是signed long或int。

signed long * signed long仍然是signed long，可能会遇到带符号整数溢出，即UB。您signed long的范围小于0x12345678 * 0x12345678的数学乘积。通过使用ULL后缀，数学运算至少使用unsigned long long数学。 @BLUEPIXY

printf("%llu\n", 0x12345678ULL * 0x12345678);
// or if the constant can not be changed
printf("%llu\n", 1ULL * SOME_BIG_CONSTANT * SOME_BIG_CONSTANT);

迂腐：当打印可能宽于int/unsigned的整数类型时，确保最终的计算结果与说明符匹配。请考虑SOME_BIG_CONSTANT可能比unsigned long long更宽。或者离开演员阵容，并应对潜在的编译器警告。

printf("%llu\n", (unsigned long long) (1ULL * SOME_BIG_CONSTANT * SOME_BIG_CONSTANT));

另见Why write 1,000,000,000 as 1000*1000*1000 in C?
和There are reasons not to use 1000 * 1000 * 1000

为什么我会得到整数溢出，我该如何解决？

1 个答案: