C++ understanding integer overflow

The sign of pop is implementation-defined (before C++11), and INT_MIN - pop will cause an overflow if it is negative. So let's first reduce the problem to positive integers only:

if (x == INT_MIN)    // INT_MIN cannot be inverted, handle it separately
    return 0;
const int sign = (x < 0) ? -1 : 1;
x *= sign;

The overflow condition is:

10 * temp + pop > INT_MAX

After simple math, we get:

temp > (INT_MAX - pop) / 10

Here pop is always non-negative, so INT_MAX - pop cannot overflow. The final result is sign * temp. Because temp is positive, -temp cannot overflow (it could overflow if temp were negative).