Multi-dimensional (2D) array dynamically using malloc()? - c

Question

Please if someone tell me what is wrong with this code by using malloc function, I dont know what to do.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <process.h>
int main()
{
int row;
int column;
int j=0;
int **ptr;
printf("Number of rows: \n");
scanf ("%d",&row);
printf("Number of columns: \n");
scanf ("%d",&column);
There is a mistake, ???? please if someone know what to do to write
int* ptr = malloc( row * sizeof(*ptr) ); /*Allocate integer pointers for the rows */
if(ptr != NULL)
{
for(int m = 0; m < row; m++) /* Loop through each row pointer to allocate memory for columns*/
{
/* Set p[i] pointer to a block of memory for 'column' number of integers */
ptr[m] = malloc(column * sizeof **ptr); /*Here, sizeof(**p) is same as sizeof(int) */
if(ptr[m] == NULL)
{
printf("Memory allocation failed. Exiting....");
exit(1);
}
}
}
else
{
printf("Memory allocation failed. Exiting....");
exit(1);
}
for(int i=0; i<row; i++)
{
for(int j=0; j<column; j++)
{
printf("Enter values [%d] [%d]: \n",i+1,j+1 );
scanf ("%d",&ptr[i][j]);
}
}
free (ptr);
system("PAUSE");
return 0;
}
Pleas for answer because i need this code in school till 3. Jan. 2014

Answer 1

Change this line…
int* ptr = malloc( row * sizeof(*ptr) );
… to this…
ptr = malloc( row * sizeof(*ptr) );
You've already declared ptr here:
int **ptr;

Answer 2

If you feel the need to use malloc (and free) for dynamic allocation, then there are some good answers already here. But please consider that you do not need to use malloc at all in your example.
You can simply declare an automatic array of variable size as shown below. Your code will be very clear and the potential for bugs will be reduced. This feature is available in the C language since ISO C99 (earlier, in some compliers).
There can be an argument to not use variable sized array declaration if the array will be extremely large. Your example is taking user input for each array element, so that tells me you're not allocating hundreds of megabytes or anything that would blow up the stack! So this will work for you just fine. Good luck.
#include <stdio.h>
int main()
{
int rows;
int columns;
printf("Number rows: ");
scanf ("%d", &rows);
printf("Stevilo columns: ");
scanf ("%d", &columns);
int values[rows][columns]; /* <-- no need to malloc(), no need to free() */
for(int i=0; i<rows; i++) {
for(int j=0; j<columns; j++) {
printf("Enter value[%d][%d]: ",i+1,j+1 );
scanf("%d", &values[i][j]);
}
}
}

Answer 3

Instead of using sizeof (ptr) and sizeof **ptr, try to use sizeof(int*) in the first malloc, and sizeof(int) in the second malloc.

Answer 4

thank you all for so fast answers, it was very helpful, now it works without any error!
I did so:
int** ptr = (int**) malloc(row * sizeof(int)); /*Allocate integer pointers for the rows */
if(ptr != NULL)
{
for(int m = 0; m < row; m++) /* Loop through each row pointer to allocate memory for columns*/
{
/* Set p[i] pointer to a block of memory for 'column' number of integers */
ptr[m] = (int*)malloc(column * sizeof(int)); /*Here, sizeof(**p) is same as sizeof(int) */
if(ptr[m] == NULL)
{
printf("Memory allocation failed. Exiting....");
exit(1);
}
}
}
else
{
printf("Memory allocation failed. Exiting....");
exit(1);
}

Answer 5

First off your code is very messy. Your malloc call should also look like this for example:
int** ptr = (int**)malloc(row * sizeof(ptr*));
The size of a pointer will always (almost all the time unless you are on a computer made in 1982) be 4 because its a memory address aka an int wich is 4 bytes. In your for loop it should read something like:
ptr[m] = (int*)malloc(column * sizeof(ptr));