Multi-dimensional Arrays

[thinhare]

I don't really understand the way multi-dimensional arrays works with function call.

1) How to declare a function that would take a multi-dimensional array as parameter.

void fun(THIS PART);

2) How to call the function.

    main() {
        fun(THIS PART);
    }

any help would be appreciated.

--------------------------------------------

#include <stdio.h>

void fun(int (*n)[][]);

main() {
        int n[2][3] = {
                { 11, 12, 13 },
                { 21, 22, 23 }
        };

        fun(&n);
}

void fun(int (*n)[][]) {
        printf("%d\n", (*n)[1][1]);
}

[thinhare]

OK. I found an example to make it clear, the key of the solution is the size of columns should be defined when declaring the function:

#include <stdio.h>

void printArray(int array[][4]); /* declare function */

int main() {
int array[3][4] = {0,1,2,3,4,5,6,7,8,9,10,11};
printArray(array);
return 0;
}

void printArray(int array[][4]) { /* define function */
int i, j;

for(i=0 ; i<3 ; i++) {
for(j=0 ; j<4 ; j++) {
printf("%2d ", array[i][j]);
}
printf("\n");
}
}

[Salem]

http://cboard.cprogramming.com/showthread.php?t=25765

[thinhare]

salem, I looked into the link but didn't get it. have I done anything crappy?

I've kept trying for hours before asking people here.

And fortunately, I found the answer elsewhere, then posted here to share with people who may concern.

Am I wrong?

[itsme86]

From the link:

USE CODE TAGS!!!

[sand_man]

Looks like you answered your own question anyway.

[Kleid-0]

This is your original code (as a reference):

#include <stdio.h>

void fun(int (*n)[][]);

main() {
	int n[2][3] = {{ 11, 12, 13 },{ 21, 22, 23 }};
	fun(&n);
}

void fun(int (*n)[][]) {
	printf("%d\n", (*n)[1][1]);
}

int n[2][3] = {{ 11, 12, 13 },{ 21, 22, 23 }};

I really wished this worked, but I'm afraid we have to stream it like this:

int n[2][3] = {11, 12, 13, 21, 22, 23 };

You had the function like this:

void fun(int (*n)[][])

But that would be a uuhh..triarray multitwilight zone something... remember we're making this a grid type array (2 dimensional), not 3 dimensional. So it would come down to something like this:

void fun(int[][3]);

And we're also going to have to remember the 3 as a column.

And then you had:

printf("%d\n", (*n)[1][1]);

Remember that's the same thing as saying:

printf("%d\n", n[0][1][1]);

And that's wrong, because we aren't dealing with an array that's 3 dimensional, we're dealing with a two dimensional array, so the correct approach would be like this:

printf("%d\n", n[1][1]);

And that would give us 22.

Here's the final code:

#include <stdio.h>

void fun(int[][3]);

int main() {
	int n[2][3] = { 11, 12, 13, 21, 22, 23 };
	fun(n);
}

void fun(int n[][3]) {
	printf("%d\n", n[1][1]);
}

[thinhare]

Oops! I got it. Let's test it:

#include <stdio.h>

void fun(int (*n)[][]);

main() {
int n[2][3] = {{ 11, 12, 13 },{ 21, 22, 23 }};
fun(&n);
}

void fun(int (*n)[][]) {
printf("%d\n", (*n)[1][1]);
}

[thinhare]

Again

[thinhare]

Uh, now I know how to work it out.

Thanks everybody..