整数相加溢出判断

判断两个数相加overflow的代码：

int add(int a, int b)
{

    if(((a>>31) & 1) == 0 && ((b>>31) & 1) == 0)
    {

        if(a <= INT_MAX-b)
            return a+b;
        else
            return 88888;
    }

    if(((a>>31) & 1) == 1 && ((b>>31) & 1) == 1)
    {
        if(a >= INT_MIN-b)
            return a+b;
        else
            return 99999;
    }
}

ote accepte

From: http://stackoverflow.com/questions/6970802/test-whether-sum-of-two-integers-might-overflow The code you saw for testing for overflow is just bogus. For signed integers, you must test like this: if (a^b < 0) overflow=0; /* opposite signs can't overflow */ else if (a>0) overflow=(b>INT_MAX-a); else overflow=(b<INT_MIN-a); Note that the cases can be simplified a lot if one of the two numbers is a constant. For unsigned integers, you can test like this: overflow = (a+b<a); This is possible because unsigned arithmetic is defined to wrap, unlike signed arithmetic which invokes undefined behavior on overflow.