Diff Between Call By Reference And Call By Pointer

Question

Can any one tell me the exact diff between call by pointer and call by reference. Actually what is happening on both case?

Eg:

Call By Reference:

void swap(int &x, int &y)
{
   int temp;
   temp = x; /* save the value at address x */
   x = y;    /* put y into x */
   y = temp; /* put x into y */

   return;
}

swap(a, b);

Call By Pointer:

void swap(int *x, int *y)
{
   int temp;
   temp = *x; /* save the value at address x */
   *x = *y; /* put y into x */
   *y = temp; /* put x into y */

   return;
}

  swap(&a, &b);

Not a great idea to learn either C++ or Java by comparing them to each other. You'll just keep getting confused by comparing the concepts. — Ravi K Thapliyal, Jul 02 '13 at 10:30
Hello, Java Persons May Know C++ Also.. Tagged to reach some experts — Vinoth Babu, Jul 02 '13 at 10:31
@Baadshah widely off-topic, but java references are more akin to C++ pointers. — Karthik T, Jul 02 '13 at 10:31
In C++, a reference is just an alias for an object. A pointer is a type that can point to objects or be null, and supports pointer arithmetic. — juanchopanza, Jul 02 '13 at 10:32
*"Hello, Java Persons May Know C++ Also.. Tagged to reach some experts"* - Well, yeah, and the people on [movies.se] might also know something about C++. Still you wouldn't cross-post this question there, no? — Christian Rau, Jul 02 '13 at 10:49

score 22 · Accepted Answer · answered Jul 02 '13 at 10:32

22

The both cases do exactly the same.

However, the small difference is, that references are never null (and inside function you are sure, that they are referencing valid variable). On the other hand, pointers may be empty or may point to invalid place in memory, causing an AV.

answered Jul 02 '13 at 10:32

Spook

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What would be the reference of `x` in the following call: `swap(*(int*)0,y)`? – Valeri Atamaniouk Jul 02 '13 at 10:37
@ValeriAtamaniouk that would be undefined behaviour. – juanchopanza Jul 02 '13 at 10:40
@ValeriAtamaniouk What is important, it would be an undefined behavior *even before entering the function*. – Spook Jul 02 '13 at 11:08

score 4 · Answer 2 · answered Jul 02 '13 at 10:33

Semantically these calls have identical results; under the hood references are implemented with pointers.

The important difference between using references and pointers is with references you have a lot less rope to hang yourself with. References are always pointing to "something", where as pointers can point to anything. For example, it is perfectly possible to do something like this

 void swap(int *x, int *y)
 {
    int temp;
    temp = *x; /* save the value at address x */
    ++x;
    *x = *y; /* put y into x */
    *y = temp; /* put x into y */

    return;
 }

And now this code could do anything, crash, run, have monkeys fly out of your noes, anything.

When in doubt, prefer references. They are a higher, safer level of abstraction on pointers.

Valeri Atamaniouk · Answer 3 · 2013-07-02T10:43:37.467

1

In your example, there is no difference.

In C++ reference is implemented using pointers.

Edit: for those who believe reference can't be NULL:

int calc(int &arg)
{
    return arg;
}

int test(int *arg)
{
    return calc(*arg);
}

Guess, what will be the result for test(0)?

edited Jul 02 '13 at 10:43

answered Jul 02 '13 at 10:31

Valeri Atamaniouk

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... except that the pointers can be NULL and the references cannot... – trojanfoe Jul 02 '13 at 10:32
@trojanfoe Reference can point to NULL. Though it is not a common case. – Valeri Atamaniouk Jul 02 '13 at 10:34
Please show the code for that. – trojanfoe Jul 02 '13 at 10:34
Dangling references from returning a ref to a local? – daniel gratzer Jul 02 '13 at 10:35
Doesn't point to NULL. – trojanfoe Jul 02 '13 at 10:35
I don't think there is a valid way of making a reference "point to NULL". It doesn't make sense from a language point of view. – juanchopanza Jul 02 '13 at 10:36
@juanchopanza pointers and references are convertable into each other. If you have a pointer type, you can convert it into reference and back. It is wrong to assume that the reference can't be `0` in every possible case. – Valeri Atamaniouk Jul 02 '13 at 10:41
@ValeriAtamaniouk a reference has to refer to an object, anything else is undefined behaviour. – juanchopanza Jul 02 '13 at 10:45
@juanchopanza I can't recall anything in the spec that tells that `lvalue reference` must point to an object. – Valeri Atamaniouk Jul 02 '13 at 10:49

score 0 · Answer 4 · answered Jul 02 '13 at 10:34

0

Your code is C++ code. In Java there are no pointers. Both examples do the same thing, functionally there is no difference.

answered Jul 02 '13 at 10:34

moony

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score 0 · Answer 5 · answered Jul 02 '13 at 10:39

I don't know exactly why this was tagged jave as there are no pointers. However, for the concept you can use both pointers and reference interchangeably if you are working on the value attribute. for example i=4 or *i=4. The advantage of using pointers is that you can manipulate the adress itself. Sometimes you will need to modify the adress (point it to something else ...) this is when pointers are different for reference; pointers allow you to work directly on the adress, reference won't

Diff Between Call By Reference And Call By Pointer

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