Thread: dynamic array

In my code, I am allocating memory for 2 elements first, then assigning value at each memory location. when I print them I get their values.

then I allocate memoary for more 3 elements then assigning value at each memory location. when I print them I get their values.

Now I have total five elements which are stoared at five memory location.

I want to print all the five elements but when i print their value I don't get expected value

Code:

#include <stdio.h>

 #include <stdlib.h> 
  
int main() 
{     
    int i;
  
    // Dynamically allocate memory using malloc() 
      int* ptr = (int*)malloc(2 * sizeof(int)); 
  
    // Check if the memory has been successfully 
    // allocated by malloc or not 
    if (ptr != NULL) 
    { 
        // Memory has been successfully allocated 
        printf("Memory successfully allocated using malloc.\n"); 
  
    // Get the elements of the array 
       for (i = 0; i < 2; i++) 
       { 
           printf("The elements of the array are: ");
           scanf ("%d", ptr + i);            
        } 
  
        // Print the elements of the array 
        printf("The elements of the array are: "); 
        for (i = 0; i < 2; i++) 
        { 
            printf("%d, ", *(ptr  + i)); 
        } 
    } 
 
 
        
    // Dynamically allocate memory using malloc() 
      ptr = (int*)realloc(ptr, 3 * sizeof(int)); 
  
    // Check if the memory has been successfully 
    // allocated by malloc or not 
    if (ptr != NULL)
        { 
          printf("\n Enter the 3 more the elements of the array are \n");
    
           // Get the elements of the array 
           for (i = 0; i < 3; i++) 
            { 
                printf("The elements of the array are: ");
                scanf ("%d", ptr + i);            
            } 
  
            // Print the elements of the array 
            printf("\nThe elements of the array are: "); 
            for (i = 0; i < 3; i++) 
            { 
                printf("%d, ", *(ptr  + i)); 
            } 
        }
        
       // Print the elements of the array 
    printf("\nThe elements of the array are: "); 
    for (i = 0; i < 5; i++) 
    { 
      printf("%d, ", *(ptr  + i)); 
     }     
             
        
    return 0; 
}

Memory successfully allocated using malloc.
The elements of the array are: 1
The elements of the array are: 2
The elements of the array are: 1, 2,
Enter the 3 more the elements of the array are
The elements of the array are: 3
The elements of the array are: 4
The elements of the array are: 5


The elements of the array are: 3, 4, 5,
The elements of the array are: 3, 4, 5, 0, 0,

Why I don't get output 1, 2, 3, 4, 5?

Edit : int* ptr = (int*)malloc(5 * sizeof(int));

Why I don't get output 1, 2, 3, 4, 5?

Because the array has only 3 int elements..

Code:

ptr = (int*)realloc(ptr, 3 * sizeof(int));

From my manpages for realloc

The realloc() function changes the size of the memory block pointed to by ptr to size bytes.

Try 5 * sizeof(int) and make sure you don't overwrite the first two ints.

Another tip. Avoid using a pointer in realloc and returning the same pointer because if realloc fails you'll get a memory leakage:

Code:

  ptr = realloc( ptr, 5 * sizeof(int) );

Instead, use an auxiliar pointer:

Code:

  int *q;

  q = realloc( ptr, 5 * sizeof(int) );
  if ( q == NULL )
  {
    free( ptr );
    /* ... do something with this fail... */
  }

  /* Successful? Now you can update ptr! */
  ptr = q;

You can create a simple wrapper:

Code:

// reallocate memory returning 1 with successful, 0 if fails.
// update the pointer if successful only.
int reallocmem( void **p, size_t size )
{ return realloc( p, size ) != NULL; }

Usage:

Code:

  if ( reallocmem( (void **)&ptr, 5 * sizeof(int) ) )
  {
    // ... do something with reallocated buffer using ptr...
  }

> // reallocate memory returning 1 with successful, 0 if fails.
That isn't even close to working properly.

Maybe this is going a little too far:

Code:

#include <stdio.h>
#include <stdlib.h>
 
#define ReallocElems(p, nelems, msg) \
    realloc_memory((void**)&(p), (nelems) * sizeof(*(p)), msg)
 
void *realloc_memory(void **p, size_t size, const char *msg) {
    void *q = realloc(*p, size);
    if (!q) {
        if (!msg) return NULL;
        perror(msg);
        free(*p);
        exit(EXIT_FAILURE);
    }
    return *p = q;
}
 
void print(const int *a, size_t size) {
    printf("[");
    for (size_t i = 0; i < size; ++i)
        printf("%s%d", (i == 0 ? "" : ", "), a[i]);
    printf("]\n");
}
 
int main() {
    int *a = NULL;
    int size = 0;
 
    ReallocElems(a, size = 3, "memory error");
    for (int i = 0; i < size; ++i)
        a[i] = i;
 
    print(a, size); // [0,1,2]
 
    {
        int newsize = 8; 
        if (ReallocElems(a, newsize, NULL)) {
            for (int i = size; i < newsize; ++i)
                a[i] = i * 10;
            size = newsize;
        }
        else
            printf("Cannot expand array. Printing old values.\n");
    }
 
    print(a, size); // [0,1,2,40,50,60,70]
 
    free(a);
 
    return 0;
}

Y'all better hope all pointer types have the same representation.
Question 4.9

> I have allocated memory for only four variable but this program show six memory locatios
That's because you counted to 6 and not 4

If you counted to 1000, it would print 1000 consecutive addresses.

It has nothing to do with the amount of memory you allocated.

> I believe my program allocates memory for four integer variabls then adds memory for five integer variabls so there is a total memory allocation for nine variabls
Yes, that's right.