CODEWITHSUNDEEP
c++cpointerscasting
Eliza
Eliza

Reputation: 369

What is the significance of (int*) in this program?

I wrote two programs which prints out the variable the pointer p points to:

First program:

#include <stdio.h>

int main (void){

    int *p;
    int a=5, q;

    p=&a;

    q=*p;
    printf("%d", q);

}

Second program:

#include <stdio.h>

int main(void)
{

    int a=5;
    int*p;

    p= (int*)&a;

    printf("%d", *p);

return 0;
}

My question:

Both the programs print the value of a which is 5. However, the second program uses p=(int*)&a; instead of just p=&a;. Could someone please tell me the significance of (int*) casting here?

Upvotes: 3

Views: 431

Answers (5)

Kunal Saini
Kunal Saini

Reputation: 148

That is typecasting ,to tell the compiler that the value that is assigned to the pointer p that is &a is an integer type(int) but there it is of no use as &a return the address that is assigned to the variable a and it is always an integer.

Upvotes: 0

cruxion effux
cruxion effux

Reputation: 1054

Casting is a way for a programmer to tell the computer that, even though the computer thinks something is one type, we want to treat it as another type.

But here the cast is of no use as here a is an integer and thus address of a will need no cast here for integer pointer.

Upvotes: 3

Abhineet
Abhineet

Reputation: 5389

Useless use of type casting. It's like, 

int a = (int) 10 ;

Upvotes: 1

undur_gongor
undur_gongor

Reputation: 15954

It's useless.

a is an int, so &a is already a pointer to int.

Upvotes: 2

Sourav Ghosh
Sourav Ghosh

Reputation: 134326

There is no significance, rather, this cast is superfluous and not required.

In your code, a is of type int, p is of type int *. Hence, 

 p= (int*)&a;

and

p= &a;

are same and the second one here is recommended.

Upvotes: 3

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