Question

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.

Answer 1

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.