Pointer(s) Error

Im reading C: How to Program and I have got up to pointers, one of the examples shown goes with a deck of cards and shuffles then displays the 52 cards.

I am having an error in the main function with the pointers.

Code:

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

void shuffle(int [][13]);
void deal(const int [][13], const char *[], const char *[]);

main()
{

	char *suit[4] = {"Hearts","Diamonds","Clubs","Spades"};
	char *face[13] = {"Ace","Deuce","Three","Four",
			"Five","Six","Seven","Eight",
			"Nine","Ten","Jack","Queen","King"};
			
	int deck[4][13] = {0};
	
	srand(time(NULL));
	
	shuffle(deck);
	deal(deck,face,suit);
	
	return 0;

}

void shuffle(int wDeck[][13])
{

	int card, row, column;
	
	for (card=1;card<=52;card++) {
	
		row = rand()%4;
		column = rand()%13;
		
		while (wDeck[row][column] != 0) {
		
			row = rand()%4;
			column = rand()%13;
		
		}
		
		wDeck[row][column]=card;
	
	}

}

void deal(const int wDeck[][13], const char *wFace[], const char *wSuit[])
{

	int card, row, column;
	
	for (card=1;card<=52;card++)
		for (row=0;row<=3;row++)
			for (column=0;column<=12;column++)
				if (wDeck[row][column]==card)
					printf("%5s of %-8s%c",
					wFace[column], wSuit[row],
					card % 2 == 0? '\n' : '\t');

}

Is the code I am currently using, the error message is this

deck.c: In function `main':
deck.c:21: warning: passing arg 1 of `deal' from incompatible pointer type
deck.c:21: warning: passing arg 2 of `deal' from incompatible pointer type
deck.c:21: warning: passing arg 3 of `deal' from incompatible pointer type

If its too any help I am using gcc

well i got it to compile by doing this

Code:

void deal(const int [][13], const char *, const char *); // prototype

...
...
     //   deal(deck,face,suit); <- change this

deal(deck,face[0],suit[0]); // <- to this

void deal(const int wDeck[][13], const char *wFace, const char *wSuit) // <- function parameters

But i got a seg fault

your seg fault is here

Code:

printf("%5s of %-8s%c",wFace[column], wSuit[row],card % 2 == 0? '\n' : '\t');

i say forget pointers for now
go back to your book and read about how to use printf properly

First of all, this is not a good example to start learning pointers. If that's what you're doing, try to start with simple non-bloated code. Especially with code that doesn't throw warnings

Next, it's not an error but just a (mild?) warning. The problem is that deal expects its parameters to be pointers to constant values while they are not declared const in main. If you would declare them const, shuffle wouldn't work anymore, because it actually does modify those values.
A possible solution is this code, but it's not quite so easy to understand because of the use of arrays. Again, really bad example.

Code:

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

void shuffle(int [][13]);
void deal(int (*const wDeck)[13],  char *const wFace[], char *const wSuit[]);

main()
{

	char *suit[4] = {"Hearts","Diamonds","Clubs","Spades"};
	char *face[13] = {"Ace","Deuce","Three","Four",
			"Five","Six","Seven","Eight",
			"Nine","Ten","Jack","Queen","King"};
			
	int deck[4][13] = {0};
	
	srand(time(NULL));
	
	shuffle(deck);
	deal(deck,face,suit);
	
	return 0;

}

void shuffle(int wDeck[][13])
{

	int card, row, column;
	
	for (card=1;card<=52;card++) {
	
		row = rand()%4;
		column = rand()%13;
		
		while (wDeck[row][column] != 0) {
		
			row = rand()%4;
			column = rand()%13;
		
		}
		
		wDeck[row][column]=card;
	
	}

}

void deal(int (*const wDeck)[13],  char *const wFace[], char *const wSuit[])
{

	int card, row, column;
	
	for (card=1;card<=52;card++)
		for (row=0;row<=3;row++)
			for (column=0;column<=12;column++)
				if (wDeck[row][column]==card)
					printf("%5s of %-8s%c",
					wFace[column], wSuit[row],
					card % 2 == 0? '\n' : '\t');

}

I think the author confuses a pointer to a constant with a constant pointer, i.e. char * const * is a non-constant pointer to a constant pointer to non-constant chars.

edit: Personally, I keep a good distance to code that mixes pointers and arrays. It's just too error-prone (for me). So if my code isn't correct either (shouldn't throw any warnings though), some guru please correct me

That seemed to have worked Nyda thanks. sand_man, my book doesnt go deeply into prinft until the standard library is talked about however it doesnt explain much with what arguments and %f type things are allowed. Any good tutorial just on printf that you know of?

Also this was actually the last example on pointers, the rest worked pretty well.

Originally Posted by sand_man

well i got it to compile by doing this

Code:

void deal(const int [][13], const char *, const char *);
...

...

i say forget pointers for now
go back to your book and read about how to use printf properly

The example uses printf properly. It breaks because you turn an array of pointers to chars into a pointer to a string.
i.e. when deal() accesses wFace[column] it now believes wFace is a pointer to a string (to "ace") and that you are trying to access character column within that string at *(wFace +column). As soon as a column > 4 is accessed [strlen("ace")+1 ('\0')] you might get a segfault (if no valid data is at that location in memory).