Pointers and 2-D arrays

[trish]

Consider the following piece of code :

int s[4][2] = {
                   {1, 2},
                   {3, 4},
                   {5, 6},
                   {7, 8}
                  };

int * p;
int (*pp)[2];

 for(int i = 0; i < 4; i++)
 {
   p = &s[i]; //incorrect 
   p = (int *)&s[i]; //alternative 1
   pp = &s[i]; //alternative 2
 }

I am aware of the fact that in C, multidimensional arrays are stored as array of arrays & so on. If the 2-D array is split into 4 single dimensional arrays (as in the above example) and &s[i] returns the address of each of these single dimensional arrays, why can't it be simply assigned to an integer pointer ? Also, exactly what is the implication of this syntax - int (*pp)[2] ?

[CommonTater]

Hi Trish...

Have you tried ...

int *ptr;

int array[4][10];

// initialize array...

ptr = &array[3][4];

that will isolate a single element from the array.

If you want to point to an entire row... p = s[i]; should work...

[trish]

p = s[i] will work just fine & I am also aware of the syntax needed to access a particular element as pointed out. My question though, is about the statements commented out as alternatives 1 & 2 and the dreaded int (*pp)[2]...

[AndrewHunter]

All int *pp[2] is doing is creating an array of two int pointers:

#include <stdlib.h>
#include <stdio.h>

int main(void){

	int *pp[2];
	int x, y;

	x = 5;
	y = 4;

	pp[0] = &x;
	pp[1] = &y;

	x = 3;
	y = 6;

	printf("%d %d", *pp[0], *pp[1]);

	getchar();
	return (0);
}

[laserlight]

Also, exactly what is the implication of this syntax - int (*pp)[2] ?

It declares pp to be a pointer to an array of 2 ints. Contrast this with an array of two pointers to int, as shown in AndrewHunter's example.

[AndrewHunter]

Alright, with the example you posted:

...
int (*pp)[2];
..
pp = &s[i]; //alternative 2

During your loop s[0] will point to a single array carrying 2 ints. So it would be equivalent to write:

int myArray[2] = {1,2};

Now with int (*pp)[2] you can think of it as creating a pointer to an int[2] array. So to assign that pointer you would write:

int myArray[2] = {1,2};
int (*pp)[2] = &myArray;

printf("%d", (*pp)[0]); //will print 1

[trish]

It declares pp to be a pointer to an array of 2 ints. Contrast this with an array of two pointers to int, as shown in AndrewHunter's example.

Now with int (*pp)[2] you can think of it as creating a pointer to an int[2] array.

How exactly is a normal pointer any different from a pointer to an array ? Any array, even a multidimensional one, even when its sub arrays are extracted, would yield a base address...why can't a normal pointer point to it ? Additionally, going by the jargon, 'pointer to an array', why can't it point to a single dimensional array, something like this :

int a[2] = {0, 1};
int (*p)[2];

p = &a[0];

[laserlight]

How exactly is a normal pointer any different from a pointer to an array ?

A pointer to an array is a "normal pointer"

Any array, even a multidimensional one, even when its sub arrays are extracted, would yield a base address...why can't a normal pointer point to it ?

What is a "normal pointer" to you?

Additionally, going by the jargon, 'pointer to an array', why can't it point to a single dimensional array, something like this :

Let's see... a is an array of 2 ints. Therefore, a[0] is an int. So, you are trying to make p point to an int, not to an array.

[AndrewHunter]

Here is a link to a good tutorial on pointers It may help answer your questions.

[trish]

A pointer to an array is a "normal pointer"


What is a "normal pointer" to you?

I meant the human readable int * p as a normal pointer & the vague looking (int * p)[2] as the weird one