G
guthena
Write a small C program to determine whether a machine's type is
little-endian or big-endian.
little-endian or big-endian.
W
Walter Roberson
First, write a small document to describe precisely what
big-endian and little-endian -mean-. There are some systems
that are arguably both at the same time. In fact, there are some
systems that *are* both, with the "type" for any given executable
determined by a bit in one of the processor control words.
U
user923005
It's a FAQ.
F
Flash Gordon
guthena wrote, On 28/06/07 20:01:
Save yourself the trouble and use a machine where char, short, int and
long are all the same size.
Alternatively, do your own homework. Once you've posted your attempt
someone might help you with it.
Save yourself the trouble and use a machine where char, short, int and
long are all the same size.
Alternatively, do your own homework. Once you've posted your attempt
someone might help you with it.
C
Chris Hills
When does this homework have to be completed by?
E
Eric Sosman
guthena wrote On 06/28/07 15:01,:
This small C program handles other possibilities, too:
#include <stdio.h>
int main(void){int q=puts("Enter 1 for big-endian or"
" 2 for little-endian: ");fflush(0);switch(q=scanf(
"%d",&q)>0?q:0){defualt
uts("mixed-endian");break;
case 1uts("big-endian");break;defau1tuts("other"
"-endian");break;case 2uts("little-endian");break;
defaultuts("anti-endian");break;}return q?!q:q;}
Enjoy!
This small C program handles other possibilities, too:
#include <stdio.h>
int main(void){int q=puts("Enter 1 for big-endian or"
" 2 for little-endian: ");fflush(0);switch(q=scanf(
"%d",&q)>0?q:0){defualt
uts("mixed-endian");break;case 1
uts("big-endian");break;defau1tuts("other""-endian");break;case 2
uts("little-endian");break;default
uts("anti-endian");break;}return q?!q:q;}Enjoy!
O
osmium
K
Keith Thompson
guthena said:
Why?
T
Tejas Kokje
Here you go..
int main()
{
int num=1;
char *cptr;
cptr = (char *)#
if (*cptr)
printf ("little endian\n");
else
printf ("big endian\n");
return 0;
}
Which company interview question is this ? :-D
Tejas Kokje
M
Martin Ambuhl
guthena said:
No, thank you. Why are you giving out assignments?
W
William Pursell
This is only portable if you allow information
to be passed in a compile time...(eg: -DBYTE_ORDER=4321...)
#include <stdio.h>
int
main(void)
{
#if BYTE_ORDER == LITTLE_ENDIAN
fputs( "little endian\n", stdout );
#elif BYTE_ORDER == BIG_ENDIAN
fputs( "big endian\n", stdout );
#else
fputs( "unknown \n", stdout );
#endif
return 0;
}
C
Chris Dollin
guthena said:
OK, I've done that.
C
Chris Hills
To save time please put Guthena's name at the top and email it directChris Dollin said:
to his lecturer.
Further to save time can I email my homework direct to you in the
future?
Chris (The other one)
M
mark_bluemel
No
C
Chris Dollin
Chris said:
You missed my signature, yes?
No problem.
I thought you were the /other/ other one. /The/ Chris, of course, being
Chris Torek.
R
Richard Bos
Tejas Kokje said:
Congratulations. You just did his homework for him. Since it's a crappy
assignment, his teacher probably thinks he did a sterling job, he'll
probably pass the course on the back of your work, and he'll end up
holding a job as a programmer _still_ thinking that this is a good
program. Which it is not; it is flawed for several reasons. I presume
_you_ will be the one to clean up after his incompetent behind, not us?
Richard
C
Chris Hills
Chris Dollin said:
Who?
Why?
S
Spade
This most probably isn't a homework question. Its a typical question
thats asked mostly in interviews in India.
Can you tell me how it is flawed? Is it because of casting from int*
to char* ?
R
Richard Tobin
[/QUOTE]
No. To determine endianness, you have to dereference the same address
as each of two different types, so a cast is quite natural. You could
do it with a union instead.
The main theoretical flaw is the assumption (perhaps implicit in the
question, though pedants will dispute this) that a system will be
either big or little endian. Though this is usually true, there have
been real machines in which it wasn't. A more thorough test would use
a value that put different values in each byte, and verify that the
were in one of the expected orders.
You also forgot to include <stdio.h>.
-- Richard
No. To determine endianness, you have to dereference the same address
as each of two different types, so a cast is quite natural. You could
do it with a union instead.
The main theoretical flaw is the assumption (perhaps implicit in the
question, though pedants will dispute this) that a system will be
either big or little endian. Though this is usually true, there have
been real machines in which it wasn't. A more thorough test would use
a value that put different values in each byte, and verify that the
were in one of the expected orders.
You also forgot to include <stdio.h>.
-- Richard
C
Chris Dollin
Chris said: