Suppose
假设
long long b = 5*1024*1024*1024; // 5 gigs, small enough for 64 bits
printf ("%lu\n",sizeof(long long)); // prints 8 (bytes) = 64 bits
but the compiler complains:
但编译器抱怨：
warning: integer overflow in expression [-Woverflow]
Why does it overflow, what am I missing?
为什么会溢出，我错过了什么？
1 个解决方案
#1
19
Because the numbers on the right hand side are of type int, not long long, so int arithmetic is performed in the expression, leading to an overflow.
因为右侧的数字是int类型，而不是很长，所以在表达式中执行int算术，导致溢出。
If you add LL to one of them, it'll promote them all.
如果你向其中一个添加LL，它将全部提升它们。
#1
19
Because the numbers on the right hand side are of type int, not long long, so int arithmetic is performed in the expression, leading to an overflow.
因为右侧的数字是int类型，而不是很长，所以在表达式中执行int算术，导致溢出。
If you add LL to one of them, it'll promote them all.
如果你向其中一个添加LL，它将全部提升它们。