explain this program please , thanks in advance

Question

#include<stdio.h>

int main()
{
int *ptr1,*ptr2;

ptr1 = (int *)1000;
ptr2 = (int *)2000;

if(ptr2 > ptr1)
   printf("Ptr2 is far from ptr1");

return(0);
}

 meaning of these below line and what if we dont include them ?

ptr1 = (int *)1000;        ptr2 = (int *)2000;

Comments

Pointer ptr1 and ptr2 are of type integers i.e ptr1 is pointing to an address that holds 4 bytes (assuming int is 4 bytes)
 

Now when you are assigning an address to the pointer you need to typecast it with reference to the pointer type .

Thus if ptr1 is pointing to 1000 then ptr+1 will point to 1004 and same for ptr 2 .

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thnks bhai

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ptr1 = (int *)1000;

it means convert 1000 to an integer address that can be legally pointed by integer pointer..

now, no matter whether integer is present at an address '1000' or not, ptr1 will be pointing to address '1000' in memory,

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thanks in turbo c if i not type case it it is still working fine  why

Answer 1

(int *)1000 ---> Here we typecast 1000 address to ineger to pointer.

we can't directly assign the address to a pointer variable.

for example in linked list when malloc function through address then we typecast it into ( struct node *)