Question

I am experimenting with pointers in C. I wrote this code.

int i1;
int i2 = 50;

int *intptr1;
int *intptr2;

intptr1 = &i1;
intptr2 = &i2;

intptr1 = intptr2;
printf("\nvalue of intptr1: %d\n", *intptr1); /*prints 50*/
dubptr1 = (double *)intptr1;
printf("\nvalue of dubptr1: %f\n", *dubptr1); /*prints 0.0000*/

I would have expected the last statement to print 50.0000. Is it actually supposed to print 0.0000 or did I make a mistake somewhere and if I did please help me understand where?

Answer 1

No, it won't necessarily print 0.000 it can even print some garbage values also in different compilers. Here, the mistake is the value is in int data type which is of 4 bytes whereas you are converting that address into double which is 8 bytes. So, double pointer is now pointing to 8 bytes so extra 4 bytes of memory can be anything due to which it is reading 8 bytes of memory in double precision floating point representation and providing different answer. Instead you can convert the value rather than converting the address data types