How to send array as pointer to function C

Question

I am trying to make a code where i send an array from a function to be erased, i do not know the length of the array, since it will be declared after a variable that will be typed by the user, so im trying to send as a pointer. but i constantly get an error. the code looks like this

int apagaarray(int *array,int l,int c){
for(int i=0;i<l;i++){
    for(int j=0;j<c;j++){
        array[i][j]=0;
    }
}

}

but it returns an error:

error: subscripted value is neither array nor pointer nor vector|

what is going on ? is there a way to send an array to a function without having the parameters ?

check the function parameters. And howmany pointers?

Pavan Chandaka
– Pavan Chandaka

2020-06-15 01:01:14 +00:00
Commented Jun 15, 2020 at 1:01 — Pavan Chandaka
– Pavan Chandaka, Commented Jun 15, 2020 at 1:01

dreais · Accepted Answer · 2020-06-15 01:11:09Z

2

i think it would be the most correct this way:


int func(int **array, int l, int c)
{
  for (int i = 0; i < l; i++) {
    for (int j = 0; j < c; j++) {
      array[i][j] = 0;
    }
  }
}

int main(void)
{
  int l = 5, c = 5;
  int **array = malloc(sizeof(int *) * l);

  for (int i = 0; i < l; i++) {
    array[i] = malloc(sizeof(int) * c);
    for (int j = 0; j < c; j++) {
      array[i][j] = j+1;
    }
  }

  //print something
  printf("%d\n", array[3][3]);
  //empty it
  // make note that im passing **array <= a ptr of ptr
  func(array, l, c);
  return 0;
}

not my usual self to provide a code but please try to run through it and see if that answers

answered Jun 15, 2020 at 1:11

dreais

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Comments

Damiox · Accepted Answer · 2020-06-15 01:09:51Z

1

Matrices should be represented as follows: Either int **array or int *array[]

The correct function is:

int apagaarray(int **array,int l,int c){
  for(int i=0;i<l;i++){
    for(int j=0;j<c;j++){
      array[i][j]=0;
    }
  }
}

answered Jun 15, 2020 at 1:09

Damiox

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Comments

Appaji Chintimi · Accepted Answer · 2020-06-15 04:46:09Z

#include <stdio.h>

void printArray(int *arr, int rows, int cols) {
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            // Use address to refer the array element.
            printf("%d ", *(arr + (i * cols) + j));
        }
        printf("\n");
    }
}

void eraseArray(int *arr, int rows, int cols) {
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
             // Use address to refer the array element.
            *(arr + (i * cols) + j) = 0;
        }
    }
}

int main()
{
    int arr[3][3] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    printArray((int *)arr, 3, 3); // Cast array to 1-dim array
    eraseArray((int *)arr, 3, 3); // Cast array to 1-dim array
    printArray((int *)arr, 3, 3); // Cast array to 1-dim array
    return 0;
}

Cast the array to one-dimensional array and pass it.

You can change the value too, Since you are passing pointers instead of values.

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