CODEWITHSUNDEEP
c++function
Aman Deep Gautam
Aman Deep Gautam

Reputation: 8777

What are the mechanisms behind this function?

I wrote a small program:

#include <iostream>

using namespace std;

int raw(int &x) {
    cout<<x<<endl;
    cout<<&x<<endl;
}

int main () {
    int s= 1;
    int &z=s;
    raw(s);
    raw(z);
    return 0;
}

The output is(as expected):

1
0x7fff5ed36894
1
0x7fff5ed36894

It works as I expect it to be but I am curious about how this is implemented internally. Is it function overloading or something else or one of the function is a wrapper around the other function to provide user-friendliness or the compiler does casting while on its own?

Upvotes: 3

Views: 153

Answers (3)

Tony
Tony

Reputation: 1626

This is how it looks in assembler:

    int s= 1;
002044A8  mov         dword ptr [s],1  
    int &z=s;
002044AF  lea         eax,[s]  
002044B2  mov         dword ptr [z],eax  
    raw(s);
002044B5  lea         eax,[s]  
002044B8  push        eax  
002044B9  call        raw (020110Eh)  
002044BE  add         esp,4  
    raw(z);
002044C1  mov         eax,dword ptr [z]  
002044C4  push        eax  
002044C5  call        raw (020110Eh)

LEA (in lea eax,[s]) means Load Effective Address so you can see how z effectively contains a pointer to location of s.

push instructions that prepare the arguments before the function call clearly show that you get (the same) pointer as an input in both cases.

This is non-optimized code.

Upvotes: 7

Mats Petersson
Mats Petersson

Reputation: 129524

When the compiler produces code for your program, when it sees the & saying that this is a reference, it really produces a pointer variable [or something, in machine code, that resembles a pointer].

So z as such will hold the address of s.

When you call raw(s), the compiler says "Ah, so the parameter to raw is a reference which means the address of s". When you do raw(z), the compiler says "Ah, we have a reference already, so lets just pass the content of z", which since you set it to s earlier, is the same address as s.

This is exactly as it should be.

Upvotes: 5

sasha.sochka
sasha.sochka

Reputation: 14715

Internally this

int s= 1;
int &z=s;
raw(s);
raw(z);

Is optimized to this:

int s = 1;
raw(s);
raw(s);

Because after you do int &z = s; variable z will be aliased to s to end end of its lifetime. So basically it will be the same as s.

Upvotes: 3

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