@Lego: I am sorry I missed the address of operator... I wanted to type:
printf(%d, *(char *) (0x0002))
But even the above is incorrect since it is not possible to take address of
a literal in C/C++.
The best way is:
const int i=0x0002;
printf(%d, *(char *) (i));
If this print 2, then its
@saurav: your code will always print 2 irrespective of the system's
endianness!
correct thing to do is:
printf(%d, *(char *) (0x0002))
--Sundeep.
On Mon, Jun 14, 2010 at 3:02 AM, Minotauraus anike...@gmail.com wrote:
How about a pointer? :D
On Jun 13, 5:56 am, debajyotisarma
@ 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
On Mon, Jun 14, 2010 at 5:13 AM, Sundeep Singh singh.sund...@gmail.comwrote:
@saurav: your code will always print 2 irrespective of the system's
endianness!
correct thing to do is:
printf(%d, *(char *) (0x0002))
--Sundeep.
... dereferencing a very low address pointer, are you sure?
printf(%d,12424);
will give the efficient solution
if it print 1 then little indian otherwise big endian.
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Printf(%d,(char)2);
If this print 2 then lsb is 2, else if 0 then msb is 2
On Jun 13, 5:56 pm, 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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