How to point a pointer to an array

[ripper079]

Hmmm, I´ve been trying to get this work but no luck. I have declared a pointer to an int array[5] but when I assign the pointer I get a warning ( differs in levels of indirection from 'int *').

#include <stdio.h>
#define SIZE 5

int main(void)
{

	int array1[SIZE] ={5,4,3,2,1};
                int array2[10] = {1,2,3,4,5,6,7,8,9,0};
                
	int (*point) [SIZE];
                /* Should work??? */ 
	point = array1;
                /* This shouln´t work 
                    point = array2 
                    or does it ??? */
return 0;
}

I just want to create a pointer that can point to an int-array of 5 element, how do i do that???

[moi]

#include <stdio.h>
#define SIZE 5

int main(void)
{

	int array1[SIZE] ={5,4,3,2,1};
                int array2[10] = {1,2,3,4,5,6,7,8,9,0};
                
	int *point;
                /* works */ 
	point = array1;
                /* works */
                point = array2;
                /* works */
return 0;
}

[Cshot]

A pointer to an array contains the address of the array's first element. Simple as that.

point = array1;

this piece of code really expands to: point = &array[0];
Therefore, just declare it as
int *point;

And make sure you don't go out of bounds when incrementing the pointer.

[ripper079]

Hmm, thats absolutly right. Maybe I asked the question wrong??
Could you explain for me the difference between

int (*point)[SIZE];
int *point[SIZE];

I know that the second declaration does but do not have a clue on the first one???

Is it a pointer to an array of SIZE int elements????

[moi]

Originally posted by ripper079
Hmm, thats absolutly right. Maybe I asked the question wrong??
Could you explain for me the difference between

int (*point)[SIZE];
int *point[SIZE];

I know that the second declaration does but do not have a clue on the first one???

Is it a pointer to an array of SIZE int elements????

they both have meaning (in fact the same meaning i believe) but neither is what you want here. i'm sorry but i'm no good at explaining these things; but a pointer doesn't refer to an array of SIZE, a pointer merely refers to the first element; you then put what ever you want in the subscript and make sure you dont out of bounds what you're pointing too. what you typed just above me is an array of pointers.

[quzah]

int *ptr[SIZE];


This is an array of pointers to integers. The size of the array (ie: the number of pointers you have) is defined as 'SIZE'.

Read from right to left:

[SIZE] = an array of SIZE size.
ptr = named ptr
* = of pointers
int = to type 'int'

Quzah.

[Cshot]

Yup. What moi said. If you want to point to an array, just use a pointer. Because the pointer would point only to a single element which is an int. Therefore you only need to declare an int pointer.

int *point[SIZE];
This is an array of pointers. You will use it if you need to point to multiple objects. If you want that to point to every single array element you can do this:

int i;
int *point[SIZE];
int array[SIZE];

for(i = 0; i < SIZE; i++)
   point[i] = &array[i];

This will cause each pointer in your array to point to each int in the int array.

[ripper079]

k, got it. Everthing is clear now exept one thing. What does

int (*point)[SIZE];

mean????

[mithrandir]

>>int (*point)[SIZE];<<

This would be used if you were using a pointer to a function (in this case that line would be included in the function header). Remembering that a pointer to a function contains the address of the function in memory. "Pointers to functions can be passed to funcitins, returned from functions, and assigned to other function pointers" (Deitel & Deitel, 2nd Edition, C: How to Program, p291).