Is it compulsory to initialize pointers in C++?

Question

Is it compulsory to initialize t in the following code, before assigning value to t? Is the code correct?

void swap(int *x, int *y)
{
    int *t;
    *t = *x;
    *x = *y;
    *y = *t;
}

You should possibly reword your question a bit. There are some seemingly contradicting replies below, though all are correct. _Assigning_ a value to an uninitialized pointer is entirely legitimate, _dereferencing_ it is not. Your code snippet dereferences, but your question is about assignment. — Damon, Mar 23 '11 at 10:43
Side comment, there is a standard defined `swap` function (well, template), but that has different semantics from what you are doing here. Are you sure that you want to confuse users by changing the semantics? (With your implementation: `swap(p,q)` is equivalent to `std::swap( *p, *q )` but not `std::swap( p, q )` the latter would swap the pointers, rather than the contents) — David Rodríguez - dribeas, Mar 23 '11 at 11:10
i know that we can use a simple variable to store the value pointed by x , and then swapping is done . But i am just experimenting whether, can we use a pointer, instead of simple variable to store the value . — Baali, Mar 24 '11 at 05:40
and this code works fine , if i initialize the pointer to 0. — Baali, Mar 24 '11 at 05:40

Nawaz · Accepted Answer · 2011-03-23T10:41:07.603

9

You don't need pointer to begin with:

void swap(int *x,int *y)
{
    int t; //not a pointer!
    t=*x;
    *x=*y;
    *y=t;
}
int a = 10, b = 20;
swap( &a, &b); //<-----------note : Needed &

--

Or maybe, you want the following swap function:

void swap(int & x,int & y) //parameters are references now!
{
    int t; //not a pointer!
    t=x;
    x=y;
    y=t;
}
int a = 10, b = 20;
swap(a,b); //<----------- Note: Not needed & anymore!

edited Mar 23 '11 at 10:41

answered Mar 23 '11 at 10:35

Nawaz

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1

His question is not "How to swap values of two variables?". – Prasoon Saurav Mar 23 '11 at 10:41
@Prasoon: I know that. But I'm telling him the alternatives that he has and can experiment with, as other answers already have explained his case! – Nawaz Mar 23 '11 at 10:42

score 8 · Answer 2 · answered Mar 23 '11 at 10:35

is the following section of code correct?

Nopes! Your code invokes Undefined behaviour because you are trying to dereference a wild pointer.

 int *t;
 *t=*x; // bad

Try this rather

 int t; // a pointer is not needed here
 t=*x;

or this

int *t = x; // initialize the pointer

score 5 · Answer 3 · answered Mar 23 '11 at 10:35

5

That code contains undefined behavior:

int *t;
*t=*x; // where will the value be copied?

Besides that it makes no sense - you need a temporary variable to store the value, not the pointer.

int t; // not a pointer
t=*x;
*x=*y;
*y=t;

answered Mar 23 '11 at 10:35

sharptooth

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score 1 · Answer 4 · answered Mar 23 '11 at 10:35

1

It's correct for a pointer.

Only references need to be initialized upon declaration (or in a constructor for instance members).

EDIT: but you got errors in your code, you shouldn't be dereferencing your parameters (ie int *ptr = otherPtr; is fine, not int *ptr = *otherPtr;)

answered Mar 23 '11 at 10:35

jv42

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Dereferencing an uninitialized pointer is undefined behavior. – Björn Pollex Mar 23 '11 at 10:37
I was only answering the OP's question, not checking the whole block of code. The question was whether declaring a pointer without assignment is valid, which it is. – jv42 Mar 23 '11 at 10:38
@Tomalak not sure what you mean? int var, *ptr = &var; is valid. – jv42 May 04 '11 at 08:21
@jv42: Yes, but it's not assignment. – Lightness Races in Orbit May 04 '11 at 08:45

score 1 · Answer 5 · answered Mar 23 '11 at 10:38

1

If you just want to make your pointer point to already-initialized data, then you don't need to initialize it. The way you do it, though, yes, you want to use one of the malloc functions to allocate enough heap space for an integer.

The proper, efficient way, to do swapping in C/C++ is

void swap(int *x, int *y) {
    int *t = x;
    x = y;
    y = t;
}

answered Mar 23 '11 at 10:38

michel-slm

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That's how I'd do it, too. I try to never leave a pointer uninitialized, and if I already have another pointer that I want to assign to it, then I just initialize to that value anyway. – Mephane Mar 23 '11 at 13:20

score 1 · Answer 6 · answered Mar 23 '11 at 10:40

You can find the right way of doing this here

#include <stdio.h>

void swap(int *i, int *j)
{
    int t;
    t = *i;
    *i = *j;
    *j = t;
}

Basically the reason has been explained to you by sharptooth but there you will find some more details and explanations about what happens in background when you do such a swap. Hope it helps to clear your ideas.

Mahesh · Answer 7 · 2011-05-03T22:06:10.680

0

int *t;
*t=*x;

t is not pointing to any valid location to be able to dereference.

is it compulsory to initialize , before assigning value to pointer t.

Yes, initializing / assigning to point to a valid memory location. Else where would it point to. It might some point to garbage and lead to undefined behavior on dereferencing.

edited May 03 '11 at 22:06

answered Mar 23 '11 at 10:36

Mahesh

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Is it compulsory to initialize pointers in C++?

7 Answers7