@ souravsain
Don't understand your solution.
if u type convert to char how u can say that msb is in higher memory address?
i think (char) will alway give the value of the lsb.
How u r checking endian ness?
normal endian ness check program
main()
{
int i=1;
char *p=(char*)&i;
if(*p==1)
printf("Small endian");
else
printf("Big endian");
}
Here suppose int is of 2 byets.
so content at the memory location of i will be
00000000 00000001
(MSB) (LSB) in Big endian machine
00000001 00000000
(LSB) (MSB) in small endian machine
as after pointing to the 1st memory location by *p we r checking if
the value is 1 or not.
if it is 1 then small endian or else big endian.
@Minotauraus
if u r creating pointer u r creating variable .
On 6/14/10, Minotauraus <anike...@gmail.com> wrote:
> How about a pointer? :D
>
> On Jun 13, 5:56 am, debajyotisarma <sarma.debajy...@gmail.com> wrote:
>> Is it possible to check endianness of a system in C without creating
>> variable?
>> i.e. Program should not contain any variable.
>
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