Thread: Endianness of machine

hey there..

my lecturer gave me the source code and ask me to give comment for each line..

Code:

#include <stdio.h> //standard I/O header file to declare printf, scanf n other things
int main() //it tells it the return type and how many parameters there are, and what their    
                //types are
{
   unsigned int i = 1; //why should i use unsigned? if use 'long', still get the same answer?
   char *c = (char*)&i; //pointer? so.. c represent character point to i? (kinda blur here)
   if (*c)
       printf("Little endian");
   else
       printf("Big endian");
   getchar(); 
   return 0;
}

please guide me.. ^_^

Originally Posted by nimitzhunter

Have you compiled thAt code? What is printed?

compiled it already..
and the result is.. Little endian

to be sure that the program working well,
i substitute Little with Big and vice versa..

Code:

#include <stdio.h> 
int main() 
   unsigned int i = 1; 
   char *c = (char*)&i; 
   if (*c)
       printf("Big endian");
   else
       printf("Little endian");
   getchar(); 
   return 0;
}

still got the same answer... Little endian

more important here,
this line... char *c = (char*)&i;
i dont know how to explain it..

Did you re-compile the code after making the changes?

Originally Posted by Bayint Naung

Did you re-compile the code after making the changes?

yup2.. still got the same answer.. is there anything wrong?

Originally Posted by naspek

compiled it already..
and the result is.. Little endian

to be sure that the program working well,
i substitute Little with Big and vice versa..

Code:

#include <stdio.h> 
int main() 
   unsigned int i = 1; 
   char *c = (char*)&i; 
   if (*c)
       printf("Big endian");
   else
       printf("Little endian");
   getchar(); 
   return 0;
}

still got the same answer... Little endian

more important here,
this line... char *c = (char*)&i;
i dont know how to explain it..

You're missing the { after main. (you need a better compiler :-)
Look at Endianness - Wikipedia, the free encyclopedia. Come back if you still don't get it.