converting a big endian to little endian

[junky_fellow]

Hi guys,

I need to convert a big endian integer to little endian integer.
(the integer is 4 bytes in size on my implementation) . I came up with
the following code. I need your comments on this. Please suggest any
improvements that can be done.

#include <stdio.h>
int main(void)
{
int big = 0x12345678;
int little;
char *big_ptr = (char *)&big;
char *little_ptr = (char *)&little;

little_ptr[0] = big_ptr[3];
little_ptr[1] = big_ptr[2];
little_ptr[2] = big_ptr[1];
little_ptr[3] = big_ptr[0];

printf("big = 0x%x little = 0x%x\n",big,lit tle);

}

[Random832]

2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:

Random832 wrote:

>2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:

>>Anoop Saxena wrote:
z = ( (x << 16) | ( x >16) );

(It also has /definitely/ superfluous brackets

Precedence rules for bitwise operators are not widely understood.

I don't doubt it, but those aren't the ones I was talking about.

Some people might think that without the ones you were talking about it
would have been the equivalent of (z=(x<<16))|(x> >16). The confusion is
that widespread.

[Anoop Saxena]

>
That converts little-endian to/from PDP-endian, not big-endian.

S

--
Stephen Sprunk "God does not play dice." --Albert Einstein

Absolutely my mistake. I got it mixed with my response to the following
thread in another C forum and mixed the posts. I apologize.
"
hi all.

Here is an interesting problem hope i will get a solution of it.

We all no the sizeof int depends on the hardware..Somew here it is
2bytes and somewhere it is 4 bytes. Now i would like to present a
interesting problem based on this important property.

1)First you have to find out what is the size of the int in the
platform.
2) now suppose you got 2 bytes or 4 bytes... so in case of 2 bytes you
can represent a int no as int x = 0xa1a2; and in 4bytes you can
represent it as int x = a1a2a3a4; so now in case of 2 byte you have to
change the byte order i.i x = a1a2 should become x = a2a1 and in case
of 4bytes x = 0xa1a2a3a4 should become x = 0xa3a4a1a2.

hope you can provide with the solution. "

[Chris Dollin]

Random832 wrote:

2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:

>Random832 wrote:

>>2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:
Anoop Saxena wrote:
z = ( (x << 16) | ( x >16) );

(It also has /definitely/ superfluous brackets

Precedence rules for bitwise operators are not widely understood.

I don't doubt it, but those aren't the ones I was talking about.

Some people might think that without the ones you were talking about it
would have been the equivalent of (z=(x<<16))|(x> >16). The confusion is
that widespread.

I find that hard to believe, although if that's what you've seen,
that's what you've seen. (If "some" is small enough I'd put it
under "some people can misunderstand /anything/" ...)

--
Chris "Perikles triumphant" Dollin
The shortcuts are all full of people using them.

[Random832]

2006-12-15 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:

Random832 wrote:

>2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:

>>Random832 wrote:

2006-12-14 <el**********@m urdoch.hpl.hp.c om>,
Chris Dollin wrote:
Anoop Saxena wrote:
> z = ( (x << 16) | ( x >16) );
>
(It also has /definitely/ superfluous brackets

Precedence rules for bitwise operators are not widely understood.

I don't doubt it, but those aren't the ones I was talking about.

Some people might think that without the ones you were talking about it
would have been the equivalent of (z=(x<<16))|(x> >16). The confusion is
that widespread.

I find that hard to believe, although if that's what you've seen,
that's what you've seen. (If "some" is small enough I'd put it
under "some people can misunderstand /anything/" ...)

In that case it would be not so much that bitwise operators are
confusing as it is that they're _intimidating_.