CODEWITHSUNDEEP
cpointers
Ant's
Ant's

Reputation: 13811

How if(some pointer) in C works?

I'm a beginner in C and I'm trying to understand the concept of pointer arithmetic:

I have a code like this : 

#include<stdio.h>
void main(){
    int a[10];
    if(a)
        printf("%d\n",*a);
}

Which prints the address of first element in array a. That's fine. But in my printf statement I'm using the * operator to print the value.

But when I look at my if statement, I wonder how without * operator, if is working on a? I mean without * operator, how the if statement accesses the object the pointer points to?

I guess i'm clear enough about my doubt, thanks in advance.

Upvotes: 1

Views: 111

Answers (3)

cnicutar
cnicutar

Reputation: 182619

Which prints the address of first element in array a

In your code *a is equivalent with a[0]. You're not printing any address, just some uninitialized value.

EDIT as per comment:

no my question is without * operator, how the if statement accesses the object the pointer points to

In your code if (a) doesn't access the contents, it only tests the address of a - which will never evaluate to 0.

Upvotes: 3

Dan Dinu
Dan Dinu

Reputation: 33368

      if (a)

basically checks if a is a null pointer.

When you declare int a[10] you allocate a space of memory of 10 integers, starting at the address 'a'.

When you printf *a the compiler prints the first element becuase you're telling it to print the element at the address *(a + offset) which in your case offset = 0;

To convince yourself you can try doing

     int *a=null;
      if (a) 
       {
          //code here won't be executed because a points to a null reference
       }

Upvotes: 0

Johannes Egger
Johannes Egger

Reputation: 4031

In C there is no type for a boolean so the body of an if-statement is executed when the condition doesn't equal 0 (zero). And it's quite sure that a doesn't point to the address 0, so the condition evaluates to true.

Upvotes: 0

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