return value Vs reference (in assembly)

Question

After taking a look at a few questions (and answers) regarding this topic, I tried the below simple code in godbolt .

#include <iostream>


class TwoInts
{
public:
    TwoInts( ) = default;

    const int& getAByRef( ) const;

    int getAByVal( ) const;

private:
    int a;
    int b;
};

const int& TwoInts::getAByRef( ) const
{
    return a;
}

int TwoInts::getAByVal( ) const
{
    return a;
}


int main( )
{
    TwoInts ti;

    const int& num1 { ti.getAByRef( ) };
    const int num2 { ti.getAByVal( ) };

    //std::cout << num1 << ' ' << num2 << '\n';
}

Now I see different codes generated for the two member functions getAByRef and getAByVal :

TwoInts::getAByRef() const:
        mov     rax, rdi
        ret
TwoInts::getAByVal() const:
        mov     eax, DWORD PTR [rdi]
        ret

I would appreciate it if someone could explain what those two different assembly instructions are doing?

Answer 1

Each member function gets this pointer as an implicit argument. In ABI used by that particular compiler ( Itanium ABI , not to be confused with Itanium architecture), this is passed as the first argument in the rdi register and a value is returned (if it's trivial, and here it is) in the rax ( eax ) register.

In the first case, when you return a by reference, you're actually returning an address of a . a is the first member, so its address is the same as that of the object, ie this . Hence, you just set rax to rdi .

In the second case, when you return a by value, you need to do actual dereferencing. That's what DWORD PTR [rdi] is doing. DWORD PTR means that you want to fetch 4 bytes ( sizeof(int) ).

If you put some data member before a , you'll see an additional offset added to rdi .