[Solved] Bad Math with value multiplied by -1?

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Although you believe your code is similar to this:

#include <iostream>

double f()
{
    double x = 3;
    return x * -1;
}

int main()
{
    std::cout << f() << std::endl;
}

The code you have actually takes the type of x from the result of a vector’s size() – this returns a std::size_t:

#include <cstdint>
#include <iostream>

double f()
{
    std::size_t x = 3;
    return x * -1;
}

int main()
{
    std::cout << f() << std::endl;
}

What happens here is that a std::size_t is multiplied by int. The rules of promotion when one argument is signed and one unsigned say that (source: CppReference)

if the unsigned operand’s conversion rank is greater or equal to the conversion rank of the signed operand, the signed operand is converted to the unsigned operand’s type.

std::size_t is usually higher rank than int, so we are now multiplying 3 by (std::size_t)-1 – a large number. Then when we return, this large number is converted to double to match the signature of f().

The issue can be avoided a few ways:

  • Store the result of size() into a variable of signed or floating-point type, and use that to multiply.
  • Convert the result of size() using static_cast<> to a signed or floating-point type when multiplying.
  • Multiply by a floating-point constant (e.g. -1.0) – this will cause the other argument to be promoted to floating-point.

solved Bad Math with value multiplied by -1?