Which default argument is evaluated first and why?

Question

I have three functions, funt1(), funt2(), and funt3().

int funt1()
{
    cout<<"funt1 called"<<endl;
    return 10;
}

int funt2()
{
    cout<<"funt2 called"<<endl;
    return 20;
}

void funt3(int x=funt1(), int y=funt2())
{
    cout << x << y << endl;
}

My main function:

int main()
{
    funt3();
    return 0;
}

When I am calling funt3() in my main() method, why is funt1() is called first, and then funt2()?

Answer 1

It depends on your compiler. Others may call funct2() first. Neither C or C++ guarantee the order of evaluation of function arguments.

See Parameter evaluation order before a function calling in C

Answer 2

C++ standard does not define that, so it's totally compiler-specific. That said, you should never rely on an instance of undefined behaviour.

EDIT: if you really want to keep functions invocations as default parameters to reduce the numbers of parameters you have to pass each time I suggest you do the following:

void funt3(int x, int y)
{
    cout<<x<<y<<endl;
}

void funt3(int x)
{
    funt3(x, funt2());
}

void funt3()
{
    funt3(funt1());
}