2D array through pointers

Question

I want to scan a 2D array with the help of pointers and have written this code, could you tell me why the compiler gives errors? I know how to use double pointers to do the same, i was experimenting with this one.

#include<stdio.h>
#include<stdlib.h>
int main(void) {
    int i,j,n,a,b;
    int (*(*p)[])[];
    printf("\n\tEnter the size of the matrix in the form aXb\t\n");
    scanf("%dX%d",&a,&b);
    p=(int (*(*p)[b])[a])malloc(b*sizeof(int (*p)[a]));
    for(i=0;i<b;i++) {
            p[i]=(int (*p)[a])malloc(a*sizeof(int));
            printf("\t\bEnter Column %d\t\n");
            for(j=0;j<a;j++)
                    scanf("%d",&p[i][j]);
    }
    return 0;
}

What do you mean by 'this one' ? The construct (int (*(*p)[b])[a]) ?? What should that do ? My gcc seems not to like that. — Andre Holzner
– Andre Holzner, Commented Sep 18, 2010 at 9:04
determinant.c:9: error: expected ‘)’ before ‘p’ determinant.c:9: error: expected ‘)’ before ‘[’ token determinant.c:9: error: expected ‘)’ before ‘[’ token determinant.c:9: error: expected ‘;’ before ‘malloc’ determinant.c:11: error: invalid use of array with unspecified bounds determinant.c:11: error: expected ‘)’ before ‘p’ determinant.c:11: error: expected ‘)’ before ‘[’ token determinant.c:11: error: invalid use of array with unspecified bounds determinant.c:11: error: expected ‘;’ before ‘malloc’ determinant.c:14: error: invalid use of array with unspecified bounds — n0nChun
– n0nChun, Commented Sep 18, 2010 at 9:14

Pete Kirkham · Accepted Answer · 2010-09-18 09:30:15Z

1

You are using pointers to arrays, so you shouldn't index them directly, as p[i] will give *(p+i) i.e. an array following the one pointed to by p, rather than an element of p.

In C, void* will convert to any pointer type, so you don't need to cast the result of malloc. If you do put the casts in, it can mask errors, for example if you are trying to assign to a non-pointer ( such as p[i] ).

In the malloc for p, sizeof(int (*p)[a]) should either use a type or an expression, not a declaration. p is a pointer to an array of pointers to arrays of int, so the type of the elements of *p is int (*)[].

So this compiles without error or warning on gcc:

#include<stdio.h>
#include<stdlib.h>
int main ( void )
{
    int i, j, n, a, b;

    int ( * ( * p ) [] ) [];

    printf ( "\n\tEnter the size of the matrix in the form aXb\t\n" );

    scanf ( "%dX%d", &a, &b );

    p = malloc ( b * sizeof ( int ( * ) [] ) );

    for ( i = 0;i < b;i++ ) {
        ( *p ) [i] = malloc ( a * sizeof ( int ) );
        printf ( "\t\bEnter Column %d\t\n", i );
        for ( j = 0;j < a;j++ )
            scanf ( "%d", & ( * ( *p ) [i] ) [j] );
    }


    return 0;
}

However, since there is no advantage in using a pointer to an array against using a pointer to its first element but it does mean you have to dereference before taking the element, it is much easier to use the pointer to a pointer form.

answered Sep 18, 2010 at 9:30

Pete Kirkham

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1 Comment

n0nChun Over a year ago

This is a brilliant answer. Thank you! :)

pmg · Accepted Answer · 2010-09-18 10:06:30Z

Do you know what int (*(*p)[])[] is?
Try cdecl.org ... http://cdecl.ridiculousfish.com/?q=int+%28%2A%28%2Ap%29%5B%5D%29%5B%5D

To use a 1-dimensional array and pretend it's a 2-dimensional one

declare a 1-dimensional object (pointer, array, whatever)
malloc a rectangular size
compute linear addressing value based on row, column, and column size
use it
free the array

That's it

/* Oh ... and use spaces in your code */
/* They are extremely cheap now a days */
#include <assert.h>
/* instead of asserting malloc and scanf, use proper error checking */
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    int i, j, n, rows, cols;
    int *p;                                            /* 1. */

    printf("Enter the size of the matrix in the form aXb\n");
    n = scanf("%dX%d", &rows, &cols);
    assert((n == 2) && ("scanf failed"));
    p = malloc(rows * cols * sizeof *p);               /* 2. */
    assert((p != NULL) && "malloc failed");
    for (i = 0; i < rows; i++) {
            int rowindex = i * cols;                   /* 3. */
            for (j = 0; j < cols; j++) {
                    n = scanf("%d", &p[rowindex + j]); /* 3. and 4. */
                    assert((n == 1) && "scanf failed");
            }
    }
    free(p);                                           /* 5. */
    return 0;
}

pmg · Accepted Answer · 2010-09-18 12:28:43Z

0

You are unnecessarily complicating the problem of accessing array elements using pointers. Try to use a simple pointer-to-pointer p.

int **p;
...
p=malloc(a*sizeof(int *));   //create one pointer for each row of matrix
...
for(i=0;i<a;i++)
{
...
p[i]=malloc(b*sizeof(int));  //create b integers in each row of matrix
...
}

edited Sep 18, 2010 at 12:28

pmg

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answered Sep 18, 2010 at 12:18

user191776

1 Comment

pmg Over a year ago

You might want to "use" the object (rather than the type) as argument to the sizeof operator. If you change p to double **p in the future, you won't need to change the lines with mallocs p = malloc(a * sizeof *p); and p[i] = malloc(b * sizeof *p[i]);

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