How Reference in C++ Works? Why do they need memory?

Question

References as they say have same address as original variable then why do they consume memory. Refer this post Reference in place of getters? in this example if I have reference as a class member the size of class increase.

Edit 1

Here is the output I get on VS 2010 Express

The program:

class X
{
    int i;
    int &I;
public:
    X():I(i){}
}x;

class Y
{
    int i;
public:
    Y(){}
}y;

int main()
{
    int j =0;
    int &J = j;

    return 0;
}

Watch Window:

Disassembly:

int main()
{
01101390  push        ebp  
01101391  mov         ebp,esp  
01101393  sub         esp,0D8h  
01101399  push        ebx  
0110139A  push        esi  
0110139B  push        edi  
0110139C  lea         edi,[ebp-0D8h]  
011013A2  mov         ecx,36h  
011013A7  mov         eax,0CCCCCCCCh  
011013AC  rep stos    dword ptr es:[edi]  
    int j =0;
011013AE  mov         dword ptr [j],0  
    int &J = j;
011013B5  lea         eax,[j]  
011013B8  mov         dword ptr [J],eax  

    return 0;
011013BB  xor         eax,eax  
}

It looks like this is implemented as exactly as a pointer in VS at least.

They don't have the same address as the original. They refer to the original. By your reasoning, two pointers to the same address shouldn't occupy two pointer's worth of memory? — Mat, Nov 16 '13 at 07:40
One of the answer for this question (http://stackoverflow.com/questions/57483/what-are-the-differences-between-pointer-variable-and-reference-variable-in-c) states "A pointer has its own memory address and size on the stack (4 bytes on x86), whereas a reference shares the same memory address (with the original variable) but also takes up some space on the stack" — Shubham, Nov 16 '13 at 07:43
That part of the answer only refers to pointers/references "on the stack". References "on the stack" might not need storage at all (they might not "exist" at all in the generated code) if something else already refers to that object in the "same scope". — Mat, Nov 16 '13 at 07:52
@Shubham That is right. The reference is just an alias for a variable, so the address-of operator has to return the address of that variable. What adds some confusion is that they are typically implemented with pointers under the hood. But one should not think of the reference as a separate entity to the thing it refers to. — juanchopanza, Nov 16 '13 at 07:53
"so the address-of operator has to return the address of that variable" @juanchopanza Sorr, I kind of not understood what you explained. Can you give an example? — Shubham, Nov 16 '13 at 19:09
@Shubham: if you do something like `int& r = x; int* p = &r;` then `p` will point to `x` not whatever behind-the-scenes storage that the compiler might be using to implement the reference `r`. — Michael Burr, Nov 16 '13 at 19:25

score 3 · Accepted Answer · answered Nov 16 '13 at 07:46

3

References are typically implemented using pointers. However note that the standard specifically says (C++11 8.3.2/4 - "References"):

It is unspecified whether or not a reference requires storage (3.7).

answered Nov 16 '13 at 07:46

Michael Burr

333,147
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score 1 · Answer 2 · answered Nov 16 '13 at 07:40

1

It takes space to store the reference itself. The same size as a pointer in fact.

answered Nov 16 '13 at 07:40

Elliott Frisch

198,278
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How Reference in C++ Works? Why do they need memory?

2 Answers2