A very stupid question on pointers

i want to know why dont we prefix the pointer with * when using it on dynamic array.

Code:

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
void main()
{

int i;
int k=10;

int *normalpointer;
normalpointer=&k;
printf("*normalpointer = %d (prefix the pointer with * here)\n", *normalpointer);

int *ptr;  //dynamic array pointer
ptr=(int *)malloc(sizeof(int)*10);   //array size 10
printf("Dynamic array created of size 10\n");
printf("printing content without prefixing the pointer by * \n");
for(i=0;i<10;i++)
{
 ptr[i]=i+1;   //no * prefix
 printf("%d  ",ptr[i]);  // *ptr[i] is illegal why ??
}

}

I dont have any problem in coding but i just need to know whats the theory going on here.

Given a pointer p and an integer n, p[n] is equivalent to *(p + n). (... which is equivalent to *(n + p), which is equivalent to n[p].)

*ptr could be written as *(ptr+0), which is the same as writing ptr[0]

Originally Posted by max1989

i want to know why dont we prefix the pointer with * when using it on dynamic array.

Code:

printf("*normalpointer = %d (prefix the pointer with * here)\n", *normalpointer);

Here you are "dereferencing" the pointer to get to the value it points to. *normalpointer in this case returns an integer.

Code:

int *ptr;  //dynamic array pointer
...
printf("%d  ",ptr[i]);  // *ptr[i] is illegal why ??
}

In this case the pointer is dereferenced by the [] brackets. ptr[i] is the same as *(ptr+i) ... go ahead try it...

*(ptr+i) worked. THANKS

Originally Posted by max1989

*(ptr+i) worked. THANKS

*smack head on desk*

There was nothing wrong with it before. I think I heard the sound of your hair parting from a distance of six light years due to concepts flying over your head