Exception not catching overflow or underflow in C++

Question

Exception not catching if two numbers are overflowing.

The output only printing the wrong number which is negative.

Test operator*(const Test &object) const
{
    try
    {
        return Test(value * object.value);
    }
    catch (std::overflow_error &e)
    {
        cout << e.what() << endl;
    }
    catch (std::underflow_error &e)
    {
        cout << e.what() << endl;
    }
}

In main:

Test t1(533222220);
Test t2(300002222);
Test t = t1 * t2; // does not throw exception but  t1 * t2 -1416076888

What is value? Please post [mcve]. If value is an integral type, multiplication never throws exceptions. — 273K, Dec 12 '21 at 22:28
@S.M. — until recently, signed arithmetic overflow resulted in undefined behavior, so it **could** throw an exception. In practice, of course, it never did. — Pete Becker, Dec 12 '21 at 22:35
@PeteBecker I think they're still undefined, despite the guaranteed two's complement. — HolyBlackCat, Dec 12 '21 at 22:39

score 3 · Answer 1 · answered Dec 12 '21 at 23:29

3

Exception not catching if two numbers are overflowing.

Correct. No exceptions are thrown by integer arithmetic. If signed integer operation overflows, then the behaviour of the program is undefined.

answered Dec 12 '21 at 23:29

eerorika

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score 1 · Answer 2 · answered Dec 12 '21 at 23:29

Despite how it sounds like, std::overflow_error and std::underflow_error does not detect arithmetic overflow and underflow by default. In standard library, overflow_error is only used for converting bitset to number, and underflow_error is never used. Any other overflow and underflow must be checked manually.

For more about how to detect overflows, you might check: How do I detect unsigned integer multiply overflow?

score 1 · Answer 3 · answered Dec 13 '21 at 05:06

In standard C++, catch blocks only work with throw statements, and there are no throw statements in the code you have shown.

Some compilers have vendor extensions that may convert non-exception errors, like integer overflows, into C++ exceptions, but your example is clearly not a case of that happening.

Exception not catching overflow or underflow in C++

3 Answers3