CODEWITHSUNDEEP
c
James Jeffery
James Jeffery

Reputation: 12599

Confusion when Swapping Pointers

In Dennis M Richies book "C programming language" it talks about pointers, and one of the examples is swapping two pointers:

void swap(int *a, int *b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp
}

What's confusing me slightly is *a = *b. If b is already a pointer being passed in, why does it need to be dereferenced before assigning it to *a? Why wouldn't a = b work as their both pointers already?

Upvotes: 2

Views: 165

Answers (4)

Vlad from Moscow
Vlad from Moscow

Reputation: 310990

The reason for the confusing is that the function actually does not swap pointers. Let's at first assume that the function has to swap pointers and its implementation has to be look as you think

void swap(int *a, int *b)
{
    int *temp;
    temp = a;
    a = b;
    b = temp
}

The function can be called the following way

int x = 10;
int y = 20;
int *px = &x;
int *py = &y;

swap( px, py );

Parameters of a function are its local variables that are initialized by arguments. So the function call can look the following way

void swap(/*int *a, int *b*/)
{
    int *a = px;
    int *b = py;

    int *temp;
    temp = a;
    a = b;
    b = temp
}

Within the function these local variables were swapped but the original arguments were not changed. Local variables a and b got copies of values stored in px and py. Thus after exiting the function these local variables will be destroyed and at the same time px and py will not be changed.

To change the pointers the function should look the following way

void swap(int **a, int **b)
{
    int *temp;
    temp = *a;
    *a = *b;
    *b = temp
}

In this case the function swaps values pointed to by pointers int **a and int **b and these values are values stored in px and py. Of course the function must be called like

swap( &px, &py );

Now let's return to the original function

void swap(int *a, int *b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp
}

It is similar to the function that swaps pointers. However in this case it swaps objects pointed to by int *a and int *b.

Upvotes: 0

Prabhu
Prabhu

Reputation: 3541

*ptr is dereferencing of the pointer.

*a = *b; means you are putting in the value contained at location b into the location where a points to. And this is what is required in the swap function.

a = b; means you are assigning value of pointer b to pointer a. This is not what you need in swap logic.

Upvotes: 0

Himanshu
Himanshu

Reputation: 4395

a = b means you are assigning address of b to a which not the right way.

for swapping two numbers means you need to get value at that address and that value you have to swap.
so right way is to do *a = *b; i.e value at the address of b assigned to value at the address of a.

Upvotes: 2

Ganesh Kamath - 'Code Frenzy'
Ganesh Kamath - 'Code Frenzy'

Reputation: 5291

On doing a = b, both a and b will point to location pointed by b, this will defeat the purpose of passing a and b as reference; as the swap() will not be reflected after the function returns.

Upvotes: 0

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