c++ overflow while calculating an arithmetic expression

Question

#include <iostream>

int main(int argc, char* argv[]) {
    std::cout << 30000 * (5 * 30000 - 22207) / 2 + 50488389 << std::endl;

    return 0;
}

The right answer I want is 1967383389, but this program outputted -180100259 after running on my computer. It seemed it had overflowed. But why? How can I judge whether an arithmetic expression will overflow?

Answer 1

When you compile your example code, the compiler usually throws a warning identifying exactly the part of your expression that overflows

main.cpp:4:24: warning: integer overflow in expression [-Woverflow]
     std::cout << 30000 * (5 * 30000 - 22207) / 2 + 50488389 << std::endl;
                  ~~~~~~^~~~~~~~~~~~~~~~~~~~~

Live Demo

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To overcome this, use all floating point constants to do this calculation:

#include <iostream>
#include <cmath>

int main(int argc, char* argv[]) {
    std::cout << std::trunc(30000.0 * (5.0 * 30000.0 - 22207.0) / 2.0 + 50488389.0) << std::endl;
                              // ^^     ^^        ^^        ^^     ^^           ^^

    return 0;
}

Live Demo

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How can I judge whether an arithmetic expression will overflow?

You can check the std::numeric_limits to see what are the maximum values, that are available with your compiler implementation and target CPU.

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If you need the exact output as it was mentioned in your question, you can use I/O manipulators to bang the output in shape:

std::cout  << std::fixed << std::setprecision(0) 
        // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
           << std::trunc(30000.0 * (5.0 * 30000.0 - 22207.0) / 2.0 + 50488389.0) << std::endl;

Live Demo