Question

From C traps and pitfalls

If a and b are two integer variables, known to be non-negative then to
test whether a+b might overflow use:

     if ((int) ((unsigned) a + (unsigned) b) < 0 )
        complain();

I didn’t get that how comparing the sum of both integers with zero will let you know that there is an overflow?

Answer 1

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.