Unsigned Short to Unsigned Long assignment

Question

Hi,

While assigning from long to short, LSB 2 bytes is 0, where as MSB is filled with values from the func1() Algorithm values from stack. Why is this happening, why the compiler is trying to get these junk values to the MSB 2bytes?

#include <stdio.h>

unsigned short func1(void); // NB: function prototype !

int main(void)

{

     unsigned long int L = 0;

     unsigned short K = 0;

     L = func1();

      printf("%lu", L); // prints junk values

      K = L; 

      printf("%u", K);  // prints 0

     return 0;
}

unsigned short func1(void)

{

      unsigned short i = 0;

      // Algorithm Logic!!!

      return i; // returns 0
}

Answer 1

try this:

L = 0x00000000 | func1();

please format code in questions in future

Answer 2

The specifier for unsigned long is lu. That for unsigned short is hu. You invoke UB by not using the proper specifiers.

Answer 3

There are a number of problems with your code - here is a fixed version which should behave correctly.

#include <stdio.h>

unsigned short func1(void); // NB: function prototype !

int main(void)
{
  unsigned long int L = 0;
  unsigned short K = 0;

  L = func1();
  printf("%lu", L);
  K = L; 
  printf("%u", K);

  return 0;
}

unsigned short func1(void)
{
   unsigned short i = 0;

   // Algorithm Logic!!!

   return i; // returns 0
}