Thread: Problem with changing variables :pointers, functions

The real problem is that they're trying to use scanf to scan into pointers that don't actually point at anything.

Quzah.

Another problem is that *a and *b are both pointers since a and b are double pointers. So, whatever you assign to *a and *b is the ADDRESS of the variable that will be accessed with **a and **b. Therefore, when you say "*a =" and "*b =", the compiler is expecting an address after the equal signs. However, c and d are intergers, not addresses. If what you want is for **a to be the exact same value as c and for **b to be the exact same value as d, you need to say

Code:

*a = &c;
*b = &d;

That uses the address operator (&) on c and d, which tells the compiler to use the address of c and d, not the actual values of c and d. Here's kinda how the compiler is reading both *a = c and *b = d.

Code:

(pointer) = (interger)

That's where the error "assignment makes pointer from interger without a cast" is coming from. You're assigning an interger value to a pointer value without a cast.

A cast is basically somthing that turns one value into another type of value. If I wanted to use casting to fix that error (which you should note that you do NOT want to do in your particular situation) I would do this.

Code:

*a = (int*)c;
*b = (int*)d;

In that code, (int*) is the cast in both lines. And by putting it in front of c and d you are "casting" those interger values as pointers to an interger.

Again, in your situation you would NOT want to do this. Let's say that c was 20 and d was 30. If you used this code

Code:

*a = (int*)c;
*b = (int*)d;

that would mean that when you used **a, the compiler would use the value stored at the memory address 20, which, 99.99999% of the time, will be memory that has absolutely nothing to do with the program you're writing. If you changed this piece of memory, you could get all kinds of different errors with your computer. The same goes for using **b. That means you're accessing the value stored at the memory address 30, which is most likely being used by some other application or program running on your computer.

thanks very much for all the replies, I see where I made my problem, there was no reason to declare the row and col variables as pointer, the function can change them if they are declared as regular integers, I guess I was trying to make it more complicated that it really is

BTW pointers are so cool

also thank you for the warm welcome, I can tell already I will be using these boards a lot in the furture =)
~Adam

Originally Posted by jverkoey

Code:

int c, d;
printf("Enter the size of matrix 1: ");
scanf("%d %d", &c, &d);

Looks like they're pointing somewhere to me....It's the assigning that's causing the problem later on.

Yeah I misread that. I thought this:

Code:

main()
{
		int *row, *col;
 
		scan(&row, &col);
 
		 printf("%d %d, row, col);
}

Was this:

Code:

main()
{
		int *row, *col;
 
		scanf("%d %d", &row, &col);
 
		 printf("%d %d, row, col);
}

Doing too many things at once I guess. However, even if that was right, it's wrong. Because you'd be assigning pointers to local variables, which would expire when the function call ended. So either way it's wrong.

Quzah.

Originally Posted by 0rion

If you wanted to use a pointer-to-pointer as argument - which you don't need to in this case but heres how you would do it:

Code:

#include <stdio.h>

void scan(int**,int**);

int main(int argc, char **argv)
{
	int* row, *col;
	
	scan(&row, &col);

	printf("%d %d",*row,*col);

	return 0;
}	

void scan(int** a, int** b)
{
	int* c, *d;

	printf("Enter the size of matrix: ");
	scanf("%d %d",c, d);

	*a = c;
	*b = d;	
}

You only really need a pointer-to-pointer argument if you wanted to modify the value of a pointer in another function. In this case you only wanted to modify the value of an integer in another function.

Um, except you'd be wrong here, because NOW you're calling scanf into pointers which don't actually point to anything. For real this time.

Quzah.

You're missing the point. Actually, the instance, but that wasn't very punny.

Where are you storing the values you scan in? You aren't. You have pointers. But those pointers don't point to anything. Remember that pointers, like every other variable, store a value. The value for the pointers is the address of an actual variable instance. You don't have any, so you can't store an integeral value any place.

It's like trying to put your clothes away and remembering you threw the drawers of your dresser away. You can't put them anywhere. You don't have anywhere to put them.

Quzah.

Hrm I kinda get what you mean (at least I think so!) but still dunnoe why it compiles and runs properly

Could it be that it stores it in a random memory location since I didn't point the pointer to any specific address. And the *a and *b statements are pointing to that random memory address that has the pointer (c and d respectively) thus getting the values properly. But it could fail if the pointers randomly pointed to the same thing coincidentally but not in this case?

Yes. Uninitialized pointers, which is what you have, point some place randomly in memory. This means it can point to say, read only memory, or some other random spot in memory which you have no right to play with. What happens if it points to some spot in memory which is vital to the stability of your operating system, and you go screwing with the value of what's there?

Usually however, you'll just get a segmentation fault and your program will crash.

In short, your code is wrong. Will it compile? Sure, because the C language doesn't care if you point to any spot in memory. It's not the language's job to make sure you don't crash something. It's your job.

You have no actual instance of any integers in your code. Furthermore, you don't dynamicly allocate any. You just have pointers. Pointers which point to some arbitrary point in memory, which you likely don't have any business tinkering with. You're lucky if all your program does is crash.

Quzah.

Nope. That doesn't work either, as I already mentioned. There you're trying to assign the address of local variables, which are destroyed when the function ends. Try again.

Quzah.