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An Insight to References in C++

, 18 Apr 2006 CPOL
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How C++ Compiler handles References

Introduction

I choose to write about references in C++ because I feel most of the people have misconceptions about references. I got this feeling because I took many C++ interviews and I seldom get correct answers about references in C++.

What is meant by references in C++? A reference is generally thought of as an aliasing of the variable it refers to. I hate the definition of references being an alias of a variable in C++. In this article, I will try to explain that there is nothing known as aliasing in C++.

Background

Both in C and in C++, there are only two ways by which a variable can be accessed, passed, or retrieved. The two ways are: 

  1. Accessing/passing variable by value
  2. Accessing/Passing variable by address - In this case pointers will come into the picture

There is no 3rd way of accessing/passing variables. A reference variable is just another pointer variable which will take its own space in memory. The most important thing about the references is that it's a type of pointer which gets automatically dereferenced (by compiler). Hard to believe? Let's see....

A Sample C++ Code using References

Lets write a simple C++ code which will use references:

#include <iostream.h>
int main()
{
    int i = 10;   // A simple integer variable
    int &j = i;   // A Reference to the variable i
    
    j++;   // Incrementing j will increment both i and j.

    // check by printing values of i and j
    cout<<  i  <<  j  <<endl; // should print 11 11

    // Now try to print the address of both variables i and j
    cout<<  &i  <<  &j  <<endl; 
    // surprisingly both print the same address and make us feel that they are
    // alias to the same memory location. 
    // In example below we will see what is the reality
    return 0;
}

References are nothing but constant pointers in C++. A statement int &i = j; will be converted by the compiler to int *const i = &j; i.e. References are nothing but constant pointers. They need initialization because constants must be initialized and since the pointer is constant, they can't point to anything else. Let's take the same example of references in C++ and this time we will use the syntax that the compiler uses when it sees references.

A Sample C++ Code using References (Compiler Generated Syntax)

#include <iostream.h>
int main()
{
    int i = 10;   		// A simple integer variable
    int *const j = &i;   	// A Reference to the variable i
    
    (*j)++;   		// Incrementing j. Since reference variables are 
			// automatically dereferenced by compiler

    // check by printing values of i and j
    cout<<  i  <<  *j  <<endl; // should print 11 11
    // A * is appended before j because it used to be reference variable
    // and it should get automatically dereferenced.
    return 0;
}

You must be wondering why I skipped the printing of address from the above example. This needs some explanation. Since reference variables are automatically dereferenced, what will happen to a statement like cout << &j << endl;. The compiler will convert the statement into cout << &*j << endl; because the variable gets automatically dereferenced. Now &* cancels each other. They become meaningless and cout prints the value at j which is nothing but the address of i because of the statement int *const j = &i;.

So the statement cout << &i << &j << endl; becomes cout << &i << &*j << endl; which is similar to printing the address of i in both the cases. This is the reason behind the same address being displayed while we try to print normal variables as well as reference variables.

A Sample C++ Code using Reference Cascading

Here we will try to look at a complex scenario and see how references will work in cascading. Let's follow the code below:

#include <iostream.h>
int main()
{
    int i = 10; // A Simple Integer variable
    int &j = i; // A Reference to the variable
    // Now we can also create a reference to reference variable. 
    int &k = j; // A reference to a reference variable
    // Similarly we can also create another reference to the reference variable k
    int &l = k; // A reference to a reference to a reference variable.

    // Now if we increment any one of them the effect will be visible on all the
    // variables.
    // First print original values
    // The print should be 10,10,10,10
    cout<<  i  <<  ","  <<  j  <<  ","  <<  k  <<  ","  <<  l  <<endl;
    // increment variable j
    j++; 
    // The print should be 11,11,11,11
    cout<<  i  <<  ","  <<  j  <<  ","  <<  k  <<  ","  <<  l  <<endl;
    // increment variable k
    k++;
    // The print should be 12,12,12,12
    cout<<  i  <<  ","  <<  j  <<  ","  <<  k  <<  ","  <<  l  <<endl;
    // increment variable l
    l++;
    // The print should be 13,13,13,13
    cout<<  i  <<  ","  <<  j  <<  ","  <<  k  <<  ","  <<  l  <<endl;
    return 0;
}

A sample C++ Code Using Reference Cascading (Compiler Generated Syntax)

Here we will see if we won't depend upon the compiler to generate constant pointers in place of reference and auto dereferencing the constant pointer, we can achieve the same results.

#include <iostream.h>
int main()
{
    int i = 10;         // A Simple Integer variable
    int *const j = &i;     // A Reference to the variable
    // The variable j will hold the address of i

    // Now we can also create a reference to reference variable. 
    int *const k = &*j;     // A reference to a reference variable
    // The variable k will also hold the address of i because j 
    // is a reference variable and 
    // it gets auto dereferenced. After & and * cancels each other 
    // k will hold the value of
    // j which it nothing but address of i

    // Similarly we can also create another reference to the reference variable k
    int *const l = &*k;     // A reference to a reference to a reference variable.
    // The variable l will also hold address of i because k holds address of i after
    // & and * cancels each other.

    // so we have seen that all the reference variable will actually holds the same
    // variable address.

    // Now if we increment any one of them the effect will be visible on all the
    // variables.
    // First print original values. The reference variables will have * prefixed because 
    // these variables gets automatically dereferenced.

    // The print should be 10,10,10,10
    cout<<  i  <<  ","  <<  *j  <<  ","  <<  *k  <<  ","  <<  *l  <<endl;
    // increment variable j
    (*j)++; 
    // The print should be 11,11,11,11
    cout<<  i  <<  ","  <<  *j  <<  ","  <<  *k  <<  ","  <<  *l  <<endl;
    // increment variable k
    (*k)++;
    // The print should be 12,12,12,12
    cout<<  i  <<  ","  <<  *j  <<  ","  <<  *k  <<  ","  <<  *l  <<endl;
    // increment variable l
    (*l)++;
    // The print should be 13,13,13,13
    cout  <<  i  <<  ","  <<  *j  <<  ","  <<  *k  <<  ","  <<  *l  <<endl;
    return 0;
}

A Reference Takes its Own Space in Memory

We can see this by checking the size of the class which has only reference variables. The example below proofs that a C++ reference is not an alias and takes its own space into the memory.

#include <iostream.h>

class Test
{
    int &i;   // int *const i;
    int &j;   // int *const j;
    int &k;   // int *const k; 
};

int main()
{    
    // This will print 12 i.e. size of 3 pointers
    cout<<  "size of class Test = "  <<   sizeof(class Test)  <<endl;
    return 0;
}

Conclusion

I hope that this article explains everything about C++ references. However I'd like to mention that C++ standard doesn't explain how reference behaviour should be implemented by the compiler. It's up to the compiler to decide, and most of the time it is implemented as a constant pointer.

Additional Notes to Support this Article

In the discussion forums for this article, people were having concerns that References are not constant pointers but aliases. I am writing one more example to support this fact. Look carefully at the example below:

#include <iostream.h>

class A
{
public:
	virtual void print() { cout<<"A.."<<endl; }
};

class B : public A
{
public:
	virtual void print() { cout<<"B.."<<endl; }
};

class C : public B
{
public:
	virtual void print() { cout<<"C.."<<endl; }
};

int main()
{
	C c1;
	A &a1 = c1;
	a1.print(); // prints C

 	A a2 = c1;
	a2.print(); // prints A
	return 0;
}

The example using references supports the virtual mechanism, i.e. looking into the virtual pointer to get the handle to correct function pointer. The interesting thing here is how the virtual mechanism is supported by the static type which is simply an alias. Virtual mechanism is supported by dynamic information which will come into the picture only when a pointer is involved. I hope this will clarify most of the doubts.

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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About the Author

Deepak Kumar Gupta
Architect
India India
I love programming and started (in 1995)even before I got some professional education on the same. Since then I've worked on IP network stack (written IPv6 stack and next generation TCP Stack), VoIP, IP Security (IKE / IPSec).
My preferred programming language is C++ and I always explore the ways to improve the A.I based systems using enhanced algorithms and data structures.

Comments and Discussions

 
GeneralRe: Confusing article PinmemberIlya Lipovsky24-Mar-06 11:52 
GeneralRe: Confusing article PinmemberMatthias Becker2-Apr-06 9:45 
GeneralRe: Confusing article PinmemberGatik G5-Apr-06 19:56 
GeneralRe: Confusing article Pinmemberjefito12-Apr-06 7:03 
OK, back into the fray.
 
Gatik G wrote:
Is it possible to write an assembly code in which you can specify one variable which will not take any space in memory/registers ? if yes, the concept mentioned above is wrong and I aplogize for that.

 
Where in your article do you talk about registers? They are not generally considered to be part of memory space (though some architectures do map registers into memory space). Here is what you say:
 
Gatik G wrote:
A reference variable is just another pointer variable which will take its own space in memory.

This seems to imply that a reference variable has its own location in the memory space, separate from other variables. Since registers get re-used, you cannot include registers in your criterion. In general, most things go through registers at one time or another.
 
Gatik G wrote:
No body in the discussions till now being able to proof that a reference will not take its own space in the memory.

I gave you an example in a prior post, with assembly code output to prove it. Would you like another?
 
void func( int* array, int val )
{
array[4] = val; // change array member directly
int& ri = array[3]; // reference to array member
ri = val; // change array member through reference
}

 
Here is the generated assembly:
 
PUBLIC ?func@@YAXPAHH@Z ; func
_TEXT SEGMENT
_array$ = 8
_val$ = 12
?func@@YAXPAHH@Z PROC NEAR ; func
; File ref2.cpp
; Line 7
mov ecx, DWORD PTR _array$[esp-4]
mov eax, DWORD PTR _val$[esp-4]
mov DWORD PTR [ecx+16], eax
; Line 9
mov DWORD PTR [ecx+12], eax
; Line 10
ret 0
?func@@YAXPAHH@Z ENDP ; func
_TEXT ENDS
 

 
Do you need to have this explained to you, or can you figure out that nowhere does the reference 'ri' exist in a separate location in memory: the code loads the address of the array into the ecx register, and then accesses the appropriate array entry via offset from the register contents. There is no place where the variable 'ri' is stored; not in memory, not in the register.
 
Gatik G wrote:
some people say its wrong but I dont understand why.

Just because you don't understand why it's wrong doesn't make it correct.
 
Again, let's go back to your article. You say in your Introduction:
Gatik G wrote:
What is meant by references in C++? A reference is generally thought of an aliasing of the variable it refers to. I hate the definition of references being and Alias of a variable in C++. In this article I will try to explain that there is nothing known as aliasing in C++.

You go on and on about aliasing, but you never explain why you don't believe in it, or don't like them. When asked, you don't provide a definition. What is aliasing, in your mind? Why is it so hateful to you?
 
Gatik G wrote:
A reference variable is just another pointer variable which will take its own space in memory.

This is contrary to what the Standard says, and contrary to examples given to you.
 
You might also check out the book "C++ Common Knowledge", by Stephen C. Dewhurst. Chapter 5 is called "References Are Aliases, Not Pointers".
 
Gatik G wrote:
The most important thing about the references is that its a type of pointer which gets automatically dereferenced (by compiler..).

The closest thing I could find to this point of view is Stroustrup's "The C++ Programming Language" (2nd ed.), where he says "The obvious implementation of a reference is as a (constant) pointer that is dereferenced each time it is used". However there is no guarantee that there will actually be a place in memory space that holds that value, which is what the Standard says. Sometimes there, is, and as shown above, sometimes there isn't.
 
I still find your article to be confusing, and just plain incorrect with respect to what the Standard says and to what code examples show. You can't seem to be bothered to address either criticism. Why not? You don't seem to want to actually define your concept of aliasing, which is so important to you since the point of the article is to prove that references are not aliases.
 
I don't know what else to tell you -- you are to be commended for trying to contribute to the community, but this article remains incorrect and unhelpful,
 
regards,
 
Jeff
GeneralRe: Confusing article PinmemberGatik G18-Apr-06 20:34 
GeneralRe: Confusing article PinmemberKenGuru18-Apr-06 22:14 
GeneralRe: Confusing article PinmemberIftahh19-Apr-06 3:24 
GeneralRe: Confusing article PinmemberIlya Lipovsky28-Apr-06 12:54 
GeneralRe: Confusing article Pinmemberjefito1-May-06 8:32 
GeneralRe: Confusing article PinmemberIlya Lipovsky1-May-06 14:13 

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