C: passing an array (pointer) to a function

Question

I have a function I would like to have a function to receive a 2D array (H), to read a specified column (col) and to pass it to another array (b). It appears that the code below is not OK has I was expecting to get a 1 printed. Any guidance is very much appreciated.

#define nline 5
#define ncol 4

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

void count0(int **M,int col, int *a);

void main(){

    int i;     
    int **H;

    H=(int**)malloc(nline*sizeof(int*));
    for(i=0;i<nline;i++){
        H[i]=(int*)malloc(ncol*sizeof(int));
    }

    H[0][0]=8;
    H[0][1]=5;
    H[0][2]=6;
    H[0][3]=0;
    H[1][0]=7;
    H[1][1]=5;
    H[1][2]=4;
    H[1][3]=0;
    H[2][0]=5;
    H[2][1]=1;
    H[2][2]=1;
    H[2][3]=7;
    H[3][0]=0;
    H[3][1]=0;
    H[3][2]=0;
    H[3][3]=2;
    H[4][0]=1;
    H[4][1]=0;
    H[4][2]=1;
    H[4][3]=4;

    int *b;

    int col=1;

    count0(H,col,&b); 

    printf("num 0=%d\n",b[2]); getchar();

}

/////////////////////////////////////////////////////////////////////////////////

void count0(int **M,int col, int *a){

    int i;

    a=(int*)malloc(1*sizeof(int));

    for(i=0;i<nline;i++){
        a[i]=M[i][col];
        a=realloc(a,(i+2)*sizeof(int));
    }
}

Side note: do type *var = malloc(size * sizeof(*var)); it's less verbose, and you don't repeat the type three times. — Shahbaz
– Shahbaz, Commented Jul 30, 2013 at 16:35
void count0(int **M,int col, int *a){ should be void count0(int **M,int col, int **a){ — BLUEPIXY
– BLUEPIXY, Commented Jul 30, 2013 at 16:38

Some programmer dude · Accepted Answer · 2013-07-30 16:38:56Z

2

Your argument for the output b is wrong. It needs to be a pointer to a pointer (like how you pass it). Right now you should get a compiler warning about it. The function prototype should be

void count0(int **M,int col, int **a);

Then in the function you need to use the dereference operator to access a:

*a = malloc(1*sizeof(int));

Note that I don't cast the return value of malloc, there no need to and neither should you.

answered Jul 30, 2013 at 16:38

Some programmer dude

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2 Comments

user16720 Over a year ago

Thks! And if instead of a be an 1D array it was a 2D array? A triple pointer would needed to be defined in the function arguments int*** a?

Some programmer dude Over a year ago

@user16720 It all depends on how you declare the variable you pass into the function, but if you do it like H (the M argument in count0) then yes.

Grijesh Chauhan · Accepted Answer · 2013-07-30 16:47:24Z

1

As you already knows the number of rows = nline in your matrix in count0 function. So you should simply allocate all memory for array a in count function one time, you don't need to re-call realloc() function.

void count0(int **M, int col, int** a){
   (*a) = malloc(nline * sizeof(int));
   for(i = 0; i < nline; i++){ 
      (*a)[i] = M[i][col];
   }
}

Note: precedence of [] is higher then * so you need () around *a

just call this function as: count0(H, col, &b);

and free(b); in main() after printf statement to deallocate memory explicitly.

edited Jul 30, 2013 at 16:47

answered Jul 30, 2013 at 16:32

Grijesh Chauhan

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3 Comments

user16720 Over a year ago

Thks! And if instead of a be an 1D array it was a 2D array? A triple pointer would needed to be defined in the function arguments int*** a?

Grijesh Chauhan Over a year ago

@user16720 yes! and Check this answer: Allocate memory 2d array in function C

Grijesh Chauhan Over a year ago

@user16720 if you wants to allocate memory for 3d

fatsmcgee · Accepted Answer · 2013-07-30 16:43:28Z

0

You are trying to have this variable:

int * b;

Point to an newly allocated array after this call:

count0(H,col,&b);

To do this, you must change the signature of count 0 to the following:

void count0(int **M,int col, int **a);

Note that int *a is now int **a. Then in the body of count0, you must use the following lines to allocate/reallocate a:

*a=(int*)malloc(1*sizeof(int));
*a=realloc(a,(i+2)*sizeof(int));

Right now, a is a pointer variable that only exists in the scope of count0. Initially its value is &b (which right now doesn't make sense since &b is an int**, and a is an int*). You have a point to newly allocated memory, which is lost after the function. The changes I suggest will have *a, and therefore b, to point to memory that can be retrieved after the function call.

answered Jul 30, 2013 at 16:43

fatsmcgee

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2 Comments

user16720 Over a year ago

Thks! And if instead of a be an 1D array it was a 2D array? A triple pointer would needed to be defined in the function arguments int*** a?

fatsmcgee Over a year ago

It depends what you want to do! If you want to allocate space for the "rows" of this 2D array, as you do earlier with H=(int**)malloc(nline*sizeof(int*));, then yes. You would pass int *** a, and call &a = (int**)malloc(nline*sizeof(int*));. Assuming you have already done this, and you want to allocate "columns", as you do earlier in the program with H[i]=(int*)malloc(ncol*sizeof(int));, you would be able to still pass int ** a and simply call a[i] = (int*)malloc(ncol*sizeof(int)); . That is because a[i] will already reference space for a int pointer.

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