Passing multidimensional array by reference in C

Question

#include <stdio.h>

void f(int *app[][20]) {
    int i, j;
    for (i =0; i< 20; i++){
        for (j=0; j<20; j++){
            *app[i][j] = i;
        }
    }
}

int main() {
  int app[20][20];
  int i, j;
  f(&app);
    for (i =0; i< 20; i++){
        for (j=0; j<20; j++){
            printf("i %d, j%d  val= %d\n", i, j, app[i][j]);
        }
    }

  return 0;
}

What exactly am I doing wrong here? I don't get any error, but there is a segmentation fault and I don't know why.

te.c:15:5: warning: passing argument 1 of ‘f’ from incompatible pointer type
   f(&app);
     ^
te.c:3:6: note: expected ‘int * (*)[20]’ but argument is of type ‘int (*)[20][20]’
 void f(int *app[][20]) {

why are you trying to run code that does not cleanly compile? — user3629249
– user3629249, Commented Apr 30, 2015 at 20:38
I ignored it as it was a warning and tried to run. Since it was a pointer error, it was bound to be a seg. fault. that's when I asked this question. Thanks! — tandem
– tandem, Commented May 2, 2015 at 7:56

David Ranieri · Accepted Answer · 2015-04-29 10:55:58Z

7

void f(int *app[][20]) { /* a 2d array of pointers to int */

should be

void f(int app[][20]) { /* a pointer to an array of 20 int's */

or

void f(int (*app)[20]) { /* a pointer to an array of 20 int's */

*app[i][j] = i;

should be

app[i][j] = i; /* you don't need to dereference */

f(&app);

should be

f(app);

edited Apr 29, 2015 at 10:55

answered Apr 29, 2015 at 10:26

David Ranieri

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4 Comments

tandem Over a year ago

Question of curiosity: When passing a value as a pointer to a function we do f(&i) and use it as void f (int *i). Why not here?

David Ranieri Over a year ago

When you pass an int to a function using int i; f(i); you are passing the value, when you use int i; f(&i); you are passing a reference to the variable

tandem Over a year ago

Why can't we pass by reference here?

David Ranieri Over a year ago

Because an array decays into a pointer when is passed to a function, in other words: int a[][20]; decays into int (*a)[20]; (a pointer to an array of 20 int's)

Gopi · Accepted Answer · 2015-04-29 10:32:51Z

2

void f(int *app[][20])

should be

void f(int app[][20])

and the call should be like

f(app);

The changes done in the function f() is visible in main() You can access the array in function f() like app[i][j]

edited Apr 29, 2015 at 10:32

answered Apr 29, 2015 at 10:26

Gopi

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Comments

xrononautis · Accepted Answer · 2023-12-08 06:46:24Z

the following code correctly implements the definition and passing of a pointer to the 'app[][]' array to function 'f'

Notice the replacement of the 'magic' numbers with #define'd values

Notice the proper use of prototypes.
This will become very important as the projects become larger and consist of multiple files. Although when the project becomes multiple files, then each source file should have its' own header file that contains the prototypes for that source file

The code cleanly compiles and runs correctly

#include <stdio.h>

#define ROWS (20)
#define COLS (20)

void f(int app[][COLS]);


int main(void) 
{
    int app[ROWS][COLS];
    int i, j;

    f(app);
        
    for (i = 0; i < ROWS; i++)
    {
        for (j = 0; j < COLS; j++)
        {
            printf("i %d, j%d  val= %d\n", i, j, app[i][j]);
        }
    }

    return 0;
}

void f(int app[][COLS]) 
{
    int i, j;
    for (i = 0; i < ROWS; i++)
    {
        for (j = 0; j < COLS; j++)
        {
            app[i][j] = i;
        }
    }
}

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