what address of memory, after malloc()

Code:

---

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

struct student
{
  int roll_no;
};

int main()
{
 
  struct student *st = (struct student*) malloc(4 * sizeof(struct student));
 
  st->roll_no = 10;
 
  printf(" value : %d \n", (st )-> roll_no);
  printf(" address : %p \n", (st )-> roll_no);
 
  return 0;
}

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value : 10
address : 0000000A

What would be addresses of the memory after malloc() ?

Just print the pointer whose value was returned by malloc:

Code:

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printf(" address : %p \n", (void*)st);

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Quote:

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Originally Posted by laserlight

Just print the pointer whose value was returned by malloc:

Code:

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printf(" address : %p \n", (void*)st);

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I have doubt Am I printing correct value of N

Code:

---

#include <stdio.h>
#include <stdlib.h>
 
struct student
{
  int roll_no;
  struct student *N; 
};
 
int main()
{
 
  struct student *st = (struct student*) malloc(4 * sizeof(struct student));
  struct student *s1 = (struct student*) malloc(4 * sizeof(struct student));
 
  printf(" address : %p \n", (void*)st);
 
  st->roll_no = 10;
  st -> N = s1;
 
  printf(" value : %d \n", (st )-> roll_no);
  printf(" N : %p \n", (st )-> N);


 
  return 0;
}

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address : 00B10DD8
value : 10
N : 00B10E00

Explain why you have such a doubt.

> I have doubt Am I printing correct value of N
Yes you are.

Quote:

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Maybe you're thinking, the difference between DD8 and E00 is 40 bytes, but the size of the structure is 8, and I'm allocating 4 of them (32 bytes).

So the difference between the two pointers should be 32 right?

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No, such a thing would be a false assumption.


Two malloc'ed pointers are un-comparable.
You can't infer anything from comparing one malloc pointer from another.