Pointer to an Array in Objective-C

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It is most likely that you would not understand this chapter until you are through the chapter related to Pointers in Objective-C.

So assuming you have a bit understanding on pointers in Objective-C programming language, let us start: An array name is a constant pointer to the first element of the array. Therefore, in the declaration −

double balance[50];

balance is a pointer to &balance[0], which is the address of the first element of the array balance. Thus, the following program fragment assigns p the address of the first element of balance −

double *p;
double balance[10];

p = balance;

It is legal to use array names as constant pointers, and vice versa. Therefore, *(balance + 4) is a legitimate way of accessing the data at balance[4].

Once you store the address of first element in p, you can access array elements using *p, *(p+1), *(p+2) and so on. Below is the example to show all the concepts discussed above −

#import <Foundation/Foundation.h>

int main () {
   
   /* an array with 5 elements */
   double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
   double *p;
   int i;

   p = balance;
 
   /* output each array element's value */
   NSLog( @"Array values using pointer\n");
   for ( i = 0; i < 5; i++ ) {
      NSLog(@"*(p + %d) : %f\n",  i, *(p + i) );
   }

   NSLog(@"Array values using balance as address\n");
   for ( i = 0; i < 5; i++ ) {
      NSLog(@"*(balance + %d) : %f\n",  i, *(balance + i) );
   }
 
   return 0;
}

When the above code is compiled and executed, it produces the following result −

2013-09-14 01:36:57.995 demo[31469] Array values using pointer
2013-09-14 01:36:57.995 demo[31469] *(p + 0) : 1000.000000
2013-09-14 01:36:57.995 demo[31469] *(p + 1) : 2.000000
2013-09-14 01:36:57.995 demo[31469] *(p + 2) : 3.400000
2013-09-14 01:36:57.995 demo[31469] *(p + 3) : 17.000000
2013-09-14 01:36:57.995 demo[31469] *(p + 4) : 50.000000
2013-09-14 01:36:57.995 demo[31469] Array values using balance as address
2013-09-14 01:36:57.995 demo[31469] *(balance + 0) : 1000.000000
2013-09-14 01:36:57.995 demo[31469] *(balance + 1) : 2.000000
2013-09-14 01:36:57.995 demo[31469] *(balance + 2) : 3.400000
2013-09-14 01:36:57.995 demo[31469] *(balance + 3) : 17.000000
2013-09-14 01:36:57.995 demo[31469] *(balance + 4) : 50.000000

In the above example, p is a pointer to double, which means it can store address of a variable of double type. Once we have address in p, then *p will give us value available at the address stored in p, as we have shown in the above example.