Thread: Pointers

Hi,

I am studying C (and pointers) now, I found a nice tutorial site (with excercises) here. I am now trying to solve the excercise under "Pointers" chapter:

Write a C program to read through an array of any type using pointers.

Right now, I have this:

Code:

#include <stdio.h>

int main(int argc, char**argv)
{
	int arr1[10] = {1,2,3,4,5,6,7,8,9,10};
	char s[10] = {'a', 'b', 'c', 'd', 'e', 'd', 'c', 'b', 'a', 'z'};
	char *p = s;
 	while (*p != '\0')
	{
		printf("&#37;c", s[p-s]);
		p++;
	}
	
	return 0;
}

// this outputs: abcdedcbaz

It reads through a CHAR array (and displays the values), but how can I read through ANY array using 1 (the same) pointer? Or is that impossible? (I don't understand C and pointers very well yet)

Edit: Is a void* pointer the solution?

Code:

#include <stdio.h>

int main(int argc, char**argv)
{
	int arr1[10] = {1,2,3,4,5,6,7,8,9,10};
	char s[10] = {'a', 'b', 'c', 'd', 'e', 'd', 'c', 'b', 'a', 'z'};
	int i;
	char *p = arr1;
 	for(i = 0; i < 10; i++)
	{
		printf("%d", p[i]);
	}
	
	return 0;
}

this outputs: 1000200030

?? :P

By the way: Do you think it's useful to learn C, (I can code C++ pretty good right now, also OOP and stuff).

good point, thanks.
however:

Code:

printf("%d\n",(int*) p[i]);

doesnt work.
edit: the other one does though, which i should have added in my first code.

I thought type-casting and the [] operator had the same priority and evaluated from the left.
EDIT: I think I mixed it up with the grouping operator, (). type-casting is below [].

Code:

((int*) p)[i]

Should work though.

Thanks for the replies! Learned a lot again. C is more complicated than C++, I think!