Endianness of machine

[friend.05]

Code to check endianness of machine

[Michael Mair]

friend.05 schrieb:

Code to check endianness of machine

You forgot to include your code.
Please make sure that it is a compiling piece of
code and explain your exact problem with it.

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Sreekanth]

#include <stdio.h>

int main()
{
long x = 0x34333231;
char *y = (char *) &x;

if(strncmp(y,"1 234",4))
printf("Big Endian");
else
printf("little Endian");
}

[Ravi Uday]

friend.05 wrote:

Code to check endianness of machine

Its in C-FAQ

Check Q 20.9 at
http://www.eskimo.com/~scs/C-faq/top.html

- Ravi

[Keith Thompson]

"Sreekanth" <sr************ ********@gmail. com> writes:

#include <stdio.h>

int main()
{
long x = 0x34333231;
char *y = (char *) &x;

if(strncmp(y,"1 234",4))
printf("Big Endian");
else
printf("little Endian");
}

Please provide context, even if the previous post was only one line.
You *have* read <http://cfaj.freeshell. org/google/>, haven't you?

The program makes the following unwarranted and unnecessary
assumptions:

ASCII character set

sizeof(long)==4

Big-endian and little-endian are the only possibilities

Output will appear even without a new-line

It's acceptable to fall off the end of main() without returning a
value (true in C99, but still poor style)

And a minor point: "int main()" is acceptable, but the more explicit
"int main(void)" is preferred.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Sreekanth]

Keith Thompson wrote:

"Sreekanth" <sr************ ********@gmail. com> writes:

#include <stdio.h>

int main()
{
long x = 0x34333231;
char *y = (char *) &x;

if(strncmp(y,"1 234",4))
printf("Big Endian");
else
printf("little Endian");
}

Please provide context, even if the previous post was only one line.
You *have* read <http://cfaj.freeshell. org/google/>, haven't you?

The program makes the following unwarranted and unnecessary
assumptions:

ASCII character set

sizeof(long)==4

Big-endian and little-endian are the only possibilities

Output will appear even without a new-line

It's acceptable to fall off the end of main() without returning a
value (true in C99, but still poor style)

And a minor point: "int main()" is acceptable, but the more explicit
"int main(void)" is preferred.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

Sorry for the bad post. I am learning C programming been a java
programmer all my life. This is the first time I am using USENET group.
So that is not an excuse for a poor post. Anyway I will see to that it
doesnt repeat

[Ico]

friend.05 <hi**********@g mail.com> wrote:

Code to check endianness of machine

Usenet etiquette
--
:wq
^X^Cy^K^X^C^C^C ^C

[Ben Bacarisse]

On Wed, 15 Feb 2006 21:42:15 -0800, Sreekanth wrote:

#include <stdio.h>

int main()
{
long x = 0x34333231;
char *y = (char *) &x;

if(strncmp(y,"1 234",4))
printf("Big Endian");
else
printf("little Endian");
}

Rather too many assumptions here (as already pointed out). Endian-ness is
essentially the relationship between byte addressing and artimetic
significance so you can investigate it like this:

#include <stdio.h>

/* Edit this if you have an old compiler: */
typedef unsigned long long int integer;

union {
integer number;
char bytes[sizeof(integer)];
} u;

int main(avoid)
{
int b;
u.number = (integer)0;
/* assign 1 to LSB, 2 to the next and so on... *.
for (b = 0; b < sizeof(integer) ; b++)
u.number |= (integer)(b + 1) << (8 * b);
/* convert byte numbers to digits and print in address order */
for (b = 0; b < sizeof(integer) ; b++)
printf("%c", '0' + u.bytes[b]);
printf("\n");
return 0;
}

Although there n! possible results on an n-byte architecture, sanity and
engineering mean that you are unlikely to come across more than 3.

--
Ben.

[stathis gotsis]

"Keith Thompson" <ks***@mib.or g> wrote in message
news:ln******** ****@nuthaus.mi b.org...

"Sreekanth" <sr************ ********@gmail. com> writes:

#include <stdio.h>

int main()
{
long x = 0x34333231;
char *y = (char *) &x;

if(strncmp(y,"1 234",4))
printf("Big Endian");
else
printf("little Endian");
}

The program makes the following unwarranted and unnecessary
assumptions:

ASCII character set

sizeof(long)==4
I think sizeof(long)>=4 .
Big-endian and little-endian are the only possibilities

Output will appear even without a new-line

It's acceptable to fall off the end of main() without returning a
value (true in C99, but still poor style)

And a minor point: "int main()" is acceptable, but the more explicit
"int main(void)" is preferred.

Also, the OP forgot to include <string.h>

[stathis gotsis]

"Ben Bacarisse" <be********@bsb .me.uk> wrote in message
news:pa******** *************** *****@bsb.me.uk ...

Endian-ness is
essentially the relationship between byte addressing and artimetic
significance so you can investigate it like this:
Yes, allow me some comments.
#include <stdio.h>

/* Edit this if you have an old compiler: */
typedef unsigned long long int integer;

union {
integer number;
char bytes[sizeof(integer)];
} u;

int main(avoid)
Is this void or am i missing something?
{
int b;
u.number = (integer)0;
/* assign 1 to LSB, 2 to the next and so on... *.
for (b = 0; b < sizeof(integer) ; b++)
u.number |= (integer)(b + 1) << (8 * b);
/* convert byte numbers to digits and print in address order */
for (b = 0; b < sizeof(integer) ; b++)
printf("%c", '0' + u.bytes[b]);
That will not print digits if sizeof(integer) exceeds 9.
printf("\n");
return 0;
}

Although there n! possible results on an n-byte architecture, sanity and
engineering mean that you are unlikely to come across more than 3.