Does calling the constructor of an empty class actually use any memory?

Question

Suppose I have a class like

class Empty{
    Empty(int a){ cout << a; }
}

And then I invoke it using

int main(){
    Empty(2);
    return 0;
}

Will this cause any memory to be allocated on the stack for the creation of an "Empty" object? Obviously, the arguments need to be pushed onto the stack, but I don't want to incur any extra overhead. Basically I am using the constructor as a static member.

The reason I want to do this is because of templates. The actual code looks like

template <int which>
class FuncName{
    template <class T>
    FuncName(const T &value){
        if(which == 1){
            // specific behavior
        }else if(which == 2){
            // other specific behavior
        }
    }
};

which allows me to write something like

int main(){
    int a = 1;
    FuncName<1>(a);
}

so that I get to specialize one template parameter, while not having to specify the type of T. Also, I am hoping the compiler will optimize the other branches away inside the constructor. If anyone knows if this is true or how to check, that would be greatly appreciated. I assumed also that throwing templates into the situation does not change the "empty class" problem from above, is that right?

Answer 1

Quoting Stroustrup:

Why is the size of an empty class not zero? To ensure that the addresses of two different objects will be different. For the same reason, "new" always returns pointers to distinct objects. Consider:

class Empty { };

void f()
{
    Empty a, b;
    if (&a == &b) cout << "impossible: report error to compiler supplier";

    Empty* p1 = new Empty;
    Empty* p2 = new Empty;
    if (p1 == p2) cout << "impossible: report error to compiler supplier";
}   

There is an interesting rule that says that an empty base class need not be represented by a separate byte:

struct X : Empty {
    int a;
    // ...
};

void f(X* p)
{
    void* p1 = p;
    void* p2 = &p->a;
    if (p1 == p2) cout << "nice: good optimizer";
}

This optimization is safe and can be most useful. It allows a programmer to use empty classes to represent very simple concepts without overhead. Some current compilers provide this "empty base class optimization".

Answer 2

It might, it might, not, depending on circumstances. If you say:

Empty e;
Empty * ep = & e;

then obviously things have to be allocated.

Answer 3

Try it and see. Many compilers will eliminate such temporary objects when asked to optimise their output.

If the disassembly is too complex, then create two functions with different numbers of such objects and see if there is any difference in the stack locations of objects surrounding them, something like:

void empty1 ( int x )
{
    using namespace std;

    int a;
    Empty e1 ( x );
    int b;

    cout << endl;
    cout << "empty1" << endl;
    cout << hex << int ( &x ) << " " << dec << ( &x - &a ) << endl;
    cout << hex << int ( &a ) << " " << dec << ( &a - &b ) << endl;
}

and then try running that compared with an empty8 function with eight Empties created. With g++ on x86, if you do take the address of any of the empties you get a location between x and a on the stack, hence including x in the output. You can't assume that the storage for objects will end up in the same order as they are declared in the source code.