Thread: Value is not being stored at correct location

Originally Posted by stahta01

Please do not destroy values returned from malloc commands!

Use a second pointer that you then change instead!

Tim S.

I want to store value from 0 to 4 in memory allocated by malloc function

store value 0 at memory location 1
store value 1 at memory location 2
store value 2 at memory location 3
store value 3 at memory location 4
store value 4 at memory location 5

How can this be done

This line allocate space for an five varaibale of type int.

Code:

  int *x = malloc(5 * sizeof(x));

Where is the return value being destroyed by malloc function

The following is one way to write your code:

Code:

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

int main ()
{
   int *x = malloc(5 * sizeof(*x)); // 5 * sizeof(int) NOT sizeof(int *)

   // NEVER change the address that malloc() returns!!!  If you change that value,
   // then when you free(x), you will corrupt the heap!!!

  int i = 0;

  for ( i = 0; i < 5; i++)
  {
     x[i] = i;  // Set the values of the array, 0 - 4
     printf( "Value of x[%d] = %d\n", i, i);
  }

  printf("\n");

  for ( i = 0; i < 5; i++)
  {           // Cast the address x to a void * to correctly print the address
     printf("Value %d store at address %p\n", x[i] ,  (void *) (x + i));
  }

  free(x);  // Free the memory allocated!

  return 0;
}

I would also use a better name for the variable x. Single letters are usually used as the loop variable for foo(), etc...

Originally Posted by rstanley

The following is one way to write your code:
I would also use a better name for the variable x. Single letters are usually used as the loop variable for foo(), etc...

Thank you rstanley. Next time I would use better name for variable