Thread: How to allocate dynamic memory for structure in my code

How to allocate dynamic memory for structure in my code

I know how to use malloc to allocate dynamic memory

Code:

int *ptr = (int *) malloc(4 * sizeof (int));
if (ptr == NULL) {
    /* Memory could not be allocated
} else {
    /* Allocation succeeded.  Do something.  */
    free(ptr);  
    ptr = NULL; /* The pointed-to-data  must not be used again,
                   unless re-assigned by using malloc
                   again. */
}

How to allocate dynamic memory for the structure in code ?

Code:

#include <stdio.h>struct Person

 { int N; float height; }p1;

void show_person(struct Person p) {
  printf( "person %d is %f meters tall.\n", p.N, p.height );
}

void change_height(struct Person *p, float new_height) {
  p->height = new_height;
}

int main(void) {
    
    p1.N = 1;
    p1.height = 2.0;
    
   show_person(p1);
   change_height(&p1, 1.9)  ;
   show_person(p1);

  return 0;
}

Why do you want to dynamically allocate memory?

Originally Posted by gajya

I have studied "dynamically allocate memory" topic. I found we mostly allocate dynamic memory for pointer and structure.

Actually, we always dynamically allocate memory for objects, where an object is some region of memory storing a value of some type. So, an object could be of pointer type, of some struct type, of some integer type, etc.

All that you need to do therefore is to take your example code for allocating 4 int objects and just change int to struct Person and you're done (for 4 struct Person objects; of course you have to get rid of the multiplication if you want only one).

That said, note that you don't need to cast the return value of malloc, and except in the case where you're storing the return value in a pointer to void, you can write:

Code:

T *p = malloc(sizeof(*p));

So if you do change the type T, you only need to do so in one place in the statement.

Originally Posted by laserlight

That said, note that you don't need to cast the return value of malloc, and except in the case where you're storing the return value in a pointer to void, you can write:
the statement.

Code:

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

struct person
 
 { int N; float height; }p1;
 
int main(void)
{
   p1.N = 1;
   p1.height = 2.0;
   
   struct person *p;
   
   p = (struct person*) malloc(4 * sizeof(struct person));
   
   if (p == NULL)
    {
        printf("Memory not Available \n");
        exit(1);
    }
  
    else
   {
       printf("Memory allocated\n");
       printf("%d\n",  (p)->N);
   
   }
   free(p);


   return 0;


}

Memory allocated
8070808
That's what I do not expect

Err... N is just a member of the struct. It doesn't tell you how much memory has been allocated. Furthermore, you printed its value before assigning it an initial value, so you got garbage.

Originally Posted by laserlight

Err... N is just a member of the struct. It doesn't tell you how much memory has been allocated. Furthermore, you printed its value before assigning it an initial value, so you got garbage.

tell me if missing something

Code:

#include <stdio.h>#include<stdlib.h>
 
struct person
  
 { int N; float height; };
 
 void display(struct person *st)
{
  printf("N : %d\n", st -> N);
  printf("height: %f \n", st -> height);
}
 
int main(void)
{


   struct person p1;
  
    struct person *st = (struct person*) malloc(4 * sizeof(struct person));
  
     if (st == NULL)
    {
        printf("Memory not Available \n");
        exit(1);
    }
  
  p1.N = 10;
  p1.height = 5.2;
  
  display(&p1); 


   free(st);
 
   return 0;
 
}

N : 10
height: 5.200000

If you want to use st, why are you still declaring and using p1?