double astrix 2D array signature question..

Quote:

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Originally Posted by transgalactic2

i remember in previous examples in this forum

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It sounds like you are rote learning rather than actually understanding the reason behind doing that.

Quote:

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Originally Posted by transgalactic2

that inorder to use int** we need to pass &arr
so we could make changes on *arr

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That would be true if arr is an int* and you wanted to change the pointer itself from within the function such that the change is reflected in the caller. Here you just want to pass a pointer to a pointer, so why take its address?

i tried to write a program for it
but i get read accsses error
i dont know why??

Code:

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

 
int what1(int **arr, int m, int n);
int what2 (int **arr, int row, int col, int m, int n);
int main()
{
        int ** arr=(int**)malloc(6*sizeof(int*));
        *arr=(int*)malloc(5*sizeof(int));

        arr[0][0]=0;
        arr[0][1]=1;
        arr[0][2]=1;
        arr[0][3]=0;
        arr[0][4]=1;
       
    arr[1][0]=0;
        arr[1][1]=0;
        arr[1][2]=1;
        arr[1][3]=0;
        arr[1][4]=0;

        arr[2][0]=0;
        arr[2][1]=0;
        arr[2][2]=1;
        arr[2][3]=1;
        arr[2][4]=1;

 arr[3][0]=0;
        arr[3][1]=1;
        arr[3][2]=0;
        arr[3][3]=0;
        arr[3][4]=0;

 arr[4][0]=1;
        arr[4][1]=0;
        arr[4][2]=0;
        arr[4][3]=1;
        arr[4][4]=1;

 arr[5][0]=1;
        arr[5][1]=0;
        arr[5][2]=1;
        arr[5][3]=1;
        arr[5][4]=0;

what1(arr,6,5) ;
        return 0;
}

int what1(int **arr, int m, int n){
int i, j, tmp, stam=0;

for(i=0; i<m; i++)
for(j=0; j<n; j++){
  tmp = what2(arr,i,j,m,n);
  if (tmp>stam)
        stam = tmp;
}
return stam;
}

int what2 (int **arr, int row, int col, int m, int n){
int i,j,tmp, stam=0;

if (row < 0 || row >= m || col < 0 || col >= n) return 0;
if (arr[row][col] == 0) return 0;
 

arr[row][col] = 0;
for(i=-1; i<2; i++)
        for(j=-1; j<2; j++){
                if(!i && !j) continue;
tmp = 1 + what2(arr, row+i, col+j, m, n);       
if (tmp > stam)  stam = tmp;
        }
arr[row][col] = 1;

return stam;
}

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Quote:

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Originally Posted by transgalactic2

i tried to write a program for it
but i get read accsses error
i dont know why??

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You only allocated space for one pointer element to point to in your dynamically allocated array of pointers.

It does not have to be so hard. For example:

Code:

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

void print(int **numbers, size_t num_rows, size_t num_cols);

int main(void)
{
    const size_t num_rows = 4;
    const size_t num_cols = 3;

    /* For simplification, assume that malloc() never returns a null pointer. */
    int **numbers = malloc(num_rows * sizeof(*numbers));
    size_t n = 1;
    size_t i;
    for (i = 0; i < num_rows; ++i)
    {
        size_t j;
        numbers[i] = malloc(num_cols * sizeof(*numbers[i]));
        for (j = 0; j < num_cols; ++j)
        {
            numbers[i][j] = n++;
        }
    }

    print(numbers, num_rows, num_cols);

    for (i = 0; i < num_rows; ++i)
    {
        free(numbers[i]);
    }
    free(numbers);

    return 0;
}

void print(int **numbers, size_t num_rows, size_t num_cols)
{
    size_t i;
    for (i = 0; i < num_rows; ++i)
    {
        size_t j;
        for (j = 0; j < num_cols; ++j)
        {
            printf("%d ", numbers[i][j]);
        }
        puts("");
    }
}

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Quote:

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Originally Posted by laserlight

It sounds like you are rote learning rather than actually understanding the reason behind doing that.


That would be true if arr is an int* and you wanted to change the pointer itself from within the function such that the change is reflected in the caller. Here you just want to pass a pointer to a pointer, so why take its address?

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yes we changed the head of the linked list
so we use &head