Question

I found the following code in my textbook:

#include<stdio.h>

void disp( int *k)
{
printf("%d",*k);
}

int main( )
{
 int i ;
 int marks[ ] = { 55, 65, 75, 56, 78, 78, 90 } ;
 for ( i = 0 ; i <= 6 ; i++ )
 disp ( &marks[i] ) ;
 return 0;
}

 }

The code works just fine, but I have doubts regarding the logic:

1. I am sending the address of variables of the array. But in the disp function I am using a pointer variable as the argument and printing the value of the pointer. So the type of argument sent from the main function should mismatch with the argument of disp. So how does it work?

2. I tried to do the same by changing the disp function as

   void disp( int (&k))
   {
   printf("%d",*k);
   }
   

   but I am getting an error. What should I do to make it work by taking an address as an argument, i.e. void disp(int &k)?

Answer 1

When you do: –

int *k = &marks[i];

The above statement is broken into: –

int *k;   -> Integer Pointer
k = &marks[i]; --> `k` points to the address of marks[i]

So, basically, k is the an integer pointer, that points to the address of the current element in your array.

So, when you print *k, it is equivalent to: – *(&marks[i]), which dereferences the value and prints the element marks[i].

So, in the below code, you can understand, how the whole process of pointer assignment and de-referencing takes place: –

int *k;  // Declaring integer pointer.

// Actually 'k' is equal to the `&marks[i]`. 
// '*k' is just the indication of it being a pointer
k = &marks[i];      

// Dereference pointer
*k = *(&marks[i]); -->  = marks[i]

printf("%d",*k);  --> printf("%d",marks[i]);

Also, since you cannot declare variable like: –

int &k = &marks[i];

You cannot have them as parameter in your function: –

void disp(int &k);

Because, eventually the address of array element passed is stored in this variable. So, it has to be int *k.