passing 2d array of pointers to function giving error

Question

So I am passing a 3 by 3 array of float points. The function foo will allocate memory for each pointer. Here is the code.

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

void foo(float ***A);

int main(int argc, char **argv) {
    float* A[3][3];
    foo(&A);
}

void foo(float ***A) {
   int i,j;
   for(i=0;i<3;i++){
      for(j=0;j<3;j++){
        A[i][j] = malloc(2*sizeof(float));
        A[i][j][0] = 21;
      }
   }
}

Why does this does not work? It throws the following error:

C:\Users\tony\Code\MPI>gcc test.c
test.c: In function 'main':
test.c:8: warning: passing argument 1 of 'foo' from incompatible pointer type
test.c:4: note: expected 'float ***' but argument is of type 'float *** (*)[3][3]'

So If I call foo(A) instead of foo(&A) I get this error instead:

C:\Users\tony\Code\MPI>gcc test.c
test.c: In function 'main':
test.c:8: warning: passing argument 1 of 'foo' from incompatible pointer type
test.c:4: note: expected 'float ***' but argument is of type 'float * (*)[3]'

typeof should be sizeof

user3386109
– user3386109

2014-03-11 03:36:29 +00:00
Commented Mar 11, 2014 at 3:36 — user3386109
– user3386109, Commented Mar 11, 2014 at 3:36

Lee Duhem · Accepted Answer · 2014-03-11 03:54:15Z

1

You could try this one:

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

void foo(float *(*A)[3][3]);

int main(int argc, char **argv) {
    float* A[3][3];
    foo(&A);
    return 0;
}

void foo(float *(*A)[3][3]) {
    int i,j;
    for(i=0;i<3;i++){
        for(j=0;j<3;j++){
            (*A)[i][j] = malloc(2*sizeof(float));
            (*A)[i][j][0] = 21;
        }
    }
}

If you does not want to change the value of a variable itself in a function, you does not need to pass the address of that variable to this function. Therefore, this simpler version also works in this case:

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

void foo(float *A[3][3]);

int main(int argc, char **argv) {
    float* A[3][3];
    foo(A);
    return 0;
}

void foo(float *A[3][3]) {
    int i,j;
    for(i=0;i<3;i++){
        for(j=0;j<3;j++){
            A[i][j] = malloc(2*sizeof(float));
            A[i][j][0] = 21;
        }
    }
}

edited Mar 11, 2014 at 3:54

answered Mar 11, 2014 at 3:38

Lee Duhem

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9 Comments

HelloWorld123456789 Over a year ago

Why pass the address of A and complicate it?

user3386109 Over a year ago

I agree with @Rik. Just declare void foo( float *A[3][3] ); and call foo(A);

anthonybell Over a year ago

isn't *A[3][3] the same as ***A? I'm confused why these are different.

Lee Duhem Over a year ago

@RikayanBandyopadhyay Because the original post tries to do it, so I just demonstrate how to do it correctly. But yes, it is unnecessary.

Lee Duhem Over a year ago

@anthonybell No, it is not. Simply, array cannot always be treated like a pointer.

|

Sunil Bojanapally · Accepted Answer · 2014-03-11 03:31:34Z

1

If you are passing a two-dimensional array to a function:

int labels[NROWS][NCOLUMNS];
f(labels);

the function's declaration must match:

void f(int labels[][NCOLUMNS])
{ ... }

or

void f(int (*ap)[NCOLUMNS]) /* ap is a pointer to an array */
{ ... }

answered Mar 11, 2014 at 3:31

Sunil Bojanapally

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3 Comments

anthonybell Over a year ago

Isn't void f(int **ap) equivalent to void f(int ap[][])?

user3386109 Over a year ago

No the important distinction is that the declaration includes the number of columns, for example in the declaration void f(int array[][3]); the number 3 is extremely important. The reason is that the compiler needs that number to compute the address of an element. For example, to locate array[2][1] the compiler computes the address as follows address = array_base_address + ((2 * 3) + 1) * sizeof(array_element_type)). Without knowing the number of columns in the array, that address calculation is not possible.

Sunil Bojanapally Over a year ago

@user3386109 Here the pointer declaration is explicit and the called function doesn't allocate memory for the array which does not need to know the overall size leaving rows back.

HelloWorld123456789 · Accepted Answer · 2014-03-11 03:41:46Z

1

float* A[3][3]; is a 2D array of pointers.

But you are passing address of A and receiving it as float ***. So the error.

Pass it as foo(A); and change function prototype as

void foo(float* A[][3]);

Also, typeof should be sizeof.

A[i][j] = malloc(2*sizeof(float));

edited Mar 11, 2014 at 3:41

answered Mar 11, 2014 at 3:33

HelloWorld123456789

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