@yogeesh :
here is the reason why you are getting ch[0] = -1...bcozz
in your code.
#include <stdio.h>
int main()
{
union s
{
int i;
char ch[2]; // this is signed
};
}
char ch[2] is declared as signed one....so when compiler see that ch[]
is declared as signed and it has all 1's entry ...so it interpret it
as -1.
now execute this code by making ch[] as unsigned .... unsigned char[2];
you will get ch[0] as 255.
so in little endian machine.... ch[0] will be -1 or 255 (depending if
ch[] is signed or unsigned). or if ch[0] = 0 ..then it is big Endian
machine.
On 6/7/12, s yogeesh <[email protected]> wrote:
> Same endian concepts comes into picture but some 1 explain why obj.ch[0] =
> -1 not 255 ??
>
> Code -
>
> #include <stdio.h>
>
> int main()
> {
> union s
> {
> int i;
> char ch[2];
> };
>
>
> union s obj;
> obj.i=255;
> printf("%d %d %d\n",obj.i,obj.ch[0],obj.ch[1]);
>
> }
>
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