Value is not being stored at correct location

I am trying to allocate dynamic memory for five integer to store value from 0 to 4
as shown

00DE2AB4 = 0
00DE2AB8 = 1
00DE2ABC = 2
00DE2AC0 = 3
00DE2AC4 = 4

Code:

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#include<stdio.h>

#include<stdlib.h>

int main ()
{
  int *x = malloc(5 * sizeof(x));
 
  int i = 0;
  for ( i = 0; i < 5; i++)
  {
      x++;
      printf( "Value of x[%d] = %p\n", i, x );
         
  }
 
  printf("\n");
 
  for ( i = 0; i < 5; i++)
  {
      *x = 1;
      *x++;
      printf("Value %d store at address %p\n", *x ,  i);
  }
 
  return 0;
 
}

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Value of x[0] = 00022AB4
Value of x[1] = 00022AB8
Value of x[2] = 00022ABC
Value of x[3] = 00022AC0
Value of x[4] = 00022AC4


Value 902931021 store at address 00000000
Value 134222834 store at address 00000001
Value 1970037078 store at address 00000002
Value 959520869 store at address 00000003
Value 808596793 store at address 00000004



Program output show that memory has been allocated for five integer but value are not being stored at address. Can anyone point out what is reason for this ?

Please do not destroy values returned from malloc commands!

Use a second pointer that you then change instead!

Tim S.

Quote:

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Originally Posted by stahta01

Please do not destroy values returned from malloc commands!

Use a second pointer that you then change instead!

Tim S.

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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:

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  int *x = malloc(5 * sizeof(x));

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Where is the return value being destroyed by malloc function

Code:

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x++;

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The malloc return value was stored in X; changing the value of X results in the value being destroyed!

Tim S.

The following is one way to write your code:

Code:

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#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;
}

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

Quote:

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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...

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