Thread: Using a function to edit a 2D array

Hi every one, I want to write a function that can do some work on a 2D array. I read that functions don't return arrays and the best way would be to set the arguments of the function as pointers to the array I want to work on. The actual program is long and requires a lot of prerequists so I made a much shorter program showing basically the same problem;

Code:

#include <stdio.h>

void testfunction(int *array[][10]);

int main (void) {
   int testarray[10][10], i, j;
   for(i=0; i<10; i++){
       for(j=0; j<10; j++) testarray[i][j]=0;
   }
   testfunction(&testarray[][]);
   for(i=0; i<10; i++){
       for(j=0; j<10; j++) printf("%d\n", testarray[i][j]);
   }  
   return 0;
}

void testfunction(int *array[][10]){
   int a, b;
   for(a=0; a<10; a++) {
       for(b=0; b<10; b++) *array[a][b]=a*b;
   }  
}

I get this error when I compile this program;

Code:

test.c:10:28: error: expected expression before ‘]’ token

Could any one please guide me in this problem or introduce me to a better method to work on a 2D array inside a function? It would be greatly appreciated, thanks...

Originally Posted by edw211

Someone correct me if I'm wrong, but you don't need to use pointer syntax here because the name of an array itself functions as a pointer. These modifications to your code worked for me:

Code:

#include <stdio.h>

void testfunction(int array[][10]);

int main (void) {
   int testarray[10][10], i, j;
   for(i=0; i<10; i++){
       for(j=0; j<10; j++) testarray[i][j]=0;
   }
   testfunction(testarray);
   for(i=0; i<10; i++){
       for(j=0; j<10; j++) printf("%d\n", testarray[i][j]);
   }  
   return 0;
}

void testfunction(int array[][10]){
   int a, b;
   for(a=0; a<10; a++) {
       for(b=0; b<10; b++) array[a][b]=a*b;
   }  
}

My understanding of pointers is currently sub par for these forums, but I think you may have been inadvertently using a "pointer to a pointer" data type. Your function doesn't need to return the array unless there's something I'm missing because the changes made to your array inside the function will persist after control is returned to main.

I was under the impression that we cannot pass an array(specially 2D)to a function without using pointer.So can we return values also from the function being called without using pointer?

Originally Posted by edw211

My understanding of pointers is currently sub par for these forums, but I think you may have been inadvertently using a "pointer to a pointer" data type. Your function doesn't need to return the array unless there's something I'm missing because the changes made to your array inside the function will persist after control is returned to main.

Why you should not try to return arrays from functions in C....

Code:

#include <stdio.h>
int* MyFunction(int a, int b, int c)
  {  static int array[3];
     array[0] = a;
     array[1] = b;
     array[2] = c;
     return array;  } // return a pointer.


int main (void)
  { int *a1, *a2;  // int pointers

    printf("calling a1 = MyFunction(10,20,30);\t");
    a1 = MyFunction(10,20,30);
    printf("a1 has %d %d %d\n",a1[0],a1[1],a1[2]);

    printf("calling a2 = MyFunction(100,200,300);\t");
    a2 = MyFunction(100,200,300);
    printf("a2 has %d %d %d\n",a2[0],a2[1],a2[2]);

    printf("\nLooks good, except...\t"); 
    printf("a1 now has %d %d %d\n",a1[0],a1[1],a1[2]);

    getchar();
    return 0; }

Must be magic!

Originally Posted by CommonTater

Why you should not try to return arrays from functions in C....

Code:

#include <stdio.h>
int* MyFunction(int a, int b, int c)
  {  static int array[3];
     array[0] = a;
     array[1] = b;
     array[2] = c;
     return array;  } // return a pointer.


int main (void)
  { int *a1, *a2;  // int pointers

    printf("calling a1 = MyFunction(10,20,30);\t");
    a1 = MyFunction(10,20,30);
    printf("a1 has %d %d %d\n",a1[0],a1[1],a1[2]);

    printf("calling a2 = MyFunction(100,200,300);\t");
    a2 = MyFunction(100,200,300);
    printf("a2 has %d %d %d\n",a2[0],a2[1],a2[2]);

    printf("\nLooks good, except...\t"); 
    printf("a1 now has %d %d %d\n",a1[0],a1[1],a1[2]);

    getchar();
    return 0; }

Must be magic!

I ran the program but couldn't understand why is this happening......Help!!

Originally Posted by mistu4u

I ran the program but couldn't understand why is this happening......Help!!

As my buddy Quzah points out it's because you are returning pointers *not* arrays. You are merely choosing to treat those pointers as arrays. Both pointers end up pointing to the same place, so when the second function call is made it changes the array, effectively changing the values at the first pointer, giving you "magic numbers" as a result.