Hello,
What is equivalent of __unaligned attribute in GNU C?
I've searched for some time and it seems that it just does not exist...
/*
For HP and new Microsoft compilers it works this way:
long_pointer = (some unaligned address)
long_value = (* (__unaligned long *) long_pointer );
This causes the compiler to emit code that gets the value
in some slow but sure way, but avoiding exceptions
*/
No need to suggest workarounds please...
Regards
--PA
Nov 14 '05
18  10079 
On Sun, 19 Jun 2005 13:36:52 +0100, Lawrence Kirby wrote: On Sat, 18 Jun 2005 14:50:47 +0000, Thomas Carter wrote:
On Thu, 19 May 2005 19:56:17 +0100, Roger Leigh wrote:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
"Pavel A." <pa*****@NOwrit emeNO.com> writes:
What is equivalent of __unaligned attribute in GNU C?
I've searched for some time and it seems that it just does not exist...
It appears not to. That's probably because it should not be needed
unless your code is buggy.
I respectfully disagree: In networking applications, for performance
reasons (to avoid data copying) one must frequently deal with portions of
buffers that are not correctly aligned.
This is a generalisation which is the for most part wrong. Not totally
wrong but still mostly wrong.
typedef unsigned int UInt32 ;
UInt32 * x = (UInt32 *) malloc(10) ;
UInt32 * p = x + 1 ;
f(p) ;
f() is now dealing with an unaligned buffer. Lots of examples similar to
this can be found e.g. in the OpenSSL library.
That's what I meant.
Thomas Carter <T.******@nanob ots.com> writes: typedef unsigned int UInt32 ;
UInt32 * x = (UInt32 *) malloc(10) ;
UInt32 * p = x + 1 ;
f(p) ;
f() is now dealing with an unaligned buffer.
No, it isn't. Perhaps you need a refresher on how 'C' pointers work?
Cheers,
--
In order to understand recursion you must first understand recursion.
Remove /-nsp/ for email.
Thomas Carter wrote: typedef unsigned int UInt32 ;
UInt32 * x = (UInt32 *) malloc(10) ;
UInt32 * p = x + 1 ;
f(p) ;
f() is now dealing with an unaligned buffer.
No it isn't.
Go over the following carefully, testing it with your implementation.
Then go back to your elementary C text and learn how pointer arithmetic
works.
#include <stdio.h>
#include <stdlib.h>
#define MAGICNUMBER 10
int main(void)
{
unsigned int *x = malloc(MAGICNUM BER);
unsigned int *p = x + 1;
unsigned i;
if (!x) {
fprintf(stderr, "The allocation of x failed.\n"
"Bailing out ...\n");
exit(EXIT_FAILU RE);
}
printf("The buffer x is of %u unsigned ints.\n"
"Each is properly aligned.\n"
"The addresses x[i] are:\n", MAGICNUMBER);
for (i = 0; i < MAGICNUMBER; i++)
printf(" &x[%u] = %p, x+%u = %p\n",
i, (void *) &x[i], i, (void *) (x + i));
printf("\nAnd p, initialized to point at x + 1, has the\n"
"same value as the correctly aligned &x[1] above:\n"
" p = %p\n", (void *) p);
free(x);
return 0;
}
[output on my implementation]
The buffer x is of 10 unsigned ints.
Each is properly aligned.
The addresses x[i] are:
&x[0] = 20d50, x+0 = 20d50
&x[1] = 20d54, x+1 = 20d54
&x[2] = 20d58, x+2 = 20d58
&x[3] = 20d5c, x+3 = 20d5c
&x[4] = 20d60, x+4 = 20d60
&x[5] = 20d64, x+5 = 20d64
&x[6] = 20d68, x+6 = 20d68
&x[7] = 20d6c, x+7 = 20d6c
&x[8] = 20d70, x+8 = 20d70
&x[9] = 20d74, x+9 = 20d74
And p, initialized to point at x + 1, has the
same value as the correctly aligned &x[1] above:
p = 20d54
Thomas Carter <T.******@nanob ots.com> writes: On Sun, 19 Jun 2005 13:36:52 +0100, Lawrence Kirby wrote:
On Sat, 18 Jun 2005 14:50:47 +0000, Thomas Carter wrote:
On Thu, 19 May 2005 19:56:17 +0100, Roger Leigh wrote:
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
"Pavel A." <pa*****@NOwrit emeNO.com> writes:
> What is equivalent of __unaligned attribute in GNU C?
> I've searched for some time and it seems that it just does not exist...
It appears not to. That's probably because it should not be needed
unless your code is buggy.
I respectfully disagree: In networking applications, for performance
reasons (to avoid data copying) one must frequently deal with portions of
buffers that are not correctly aligned.
This is a generalisation which is the for most part wrong. Not totally
wrong but still mostly wrong.
typedef unsigned int UInt32 ;
UInt32 * x = (UInt32 *) malloc(10) ;
UInt32 * p = x + 1 ;
f(p) ;
f() is now dealing with an unaligned buffer. Lots of examples similar to
1) Wrong. p will point to x + sizeof(UInt32) .
2) If UInt32 means "unsigned int 32 bits long" there is no guarantee
that a plain "unsigned int" is 32 bits long. Actually, with my
compiler & platform of choice, it is 64 bits long.
--
Maurizio Loreti http://www.pd.infn.it/~loreti/mlo.html
Dept. of Physics, Univ. of Padova, Italy ROT13: yb****@cq.vasa. vg
In article <BS************ *****@newsread3 .news.atl.earth link.net>
Martin Ambuhl <ma*****@earthl ink.net> wrote: #define MAGICNUMBER 10
[the magic number came from elsewhere; all Martin Ambuhl did was
move it to a #define.]
unsigned int *x = malloc(MAGICNUM BER);
.... if (!x) {
fprintf(stderr, "The allocation of x failed.\n"
"Bailing out ...\n");
exit(EXIT_FAILU RE);
}
printf("The buffer x is of %u unsigned ints.\n"
"Each is properly aligned.\n"
"The addresses x[i] are:\n", MAGICNUMBER);
It is worth noting here that "x" now points to ten *bytes*, not
ten "unsigned int"s. If sizeof(unsigned int) is 2, this is five
"unsigned int"s; if sizeof(unsigned int) is 4, this is two with
two leftover bytes; and if sizeof(unsigned int) is 8, this is one
unsigned int, with two leftover bytes again. Only in the case when
sizeof(unsigned int) == 1 -- which is probably not supported by
GCC, although this does occur on some standalone implementations
-- is it ten unsigned ints.
Of course, it is true that sometimes one must deal with "misaligned "
memory, in places like network drivers (Ethernet headers are,
inconveniently, 14 bytes instead of 16). But there are ways to
deal with this without ever using a single line of nonstandard C.
Sometimes -- perhaps even "often" or "usually" -- the portability
gain of doing byte-at-a-time arithmetic to work with 16 and 32 bit
values outweighs the (usually tiny) performance gain one can obtain
using a compiler's "special features" for accessing unaligned value.
(The only way to find out for sure, of course, is to do both,
and measure.)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
Chris Torek wrote: It is worth noting here that "x" now points to ten *bytes*, not
ten "unsigned int"s.
Oh crap. Thank you.
On Sun, 19 Jun 2005 08:51:01 -0700, Paul Pluzhnikov wrote: Thomas Carter <T.******@nanob ots.com> writes:
typedef unsigned int UInt32 ;
UInt32 * x = (UInt32 *) malloc(10) ;
UInt32 * p = x + 1 ;
f(p) ;
f() is now dealing with an unaligned buffer.
No, it isn't. Perhaps you need a refresher on how 'C' pointers work?
I actually forgot the definition of f :-(
void f(UInt32 * p)
{
UInt64 * q = (UInt64 *) p ;
*q = 1234ULL ;
return ;
}
with UInt64 typedef'd in the obvious way. Isn't there an unaligned access
on q now?
Thomas Carter <T.******@nanob ots.com> writes: void f(UInt32 * p)
{
UInt64 * q = (UInt64 *) p ;
*q = 1234ULL ;
return ;
}
with UInt64 typedef'd in the obvious way. Isn't there an unaligned access
on q now?
Yes, now there is. However, that code is now wrong, or at least
assumes that the caller passes a pointer to UInt64 without requiring
the caller to do so (which is a bad idea(TM)).
You claimed that such code is common (it isn't and it's obviously
broken) and that you've found such code in OpenSSL (did you?).
Cheers,
--
In order to understand recursion you must first understand recursion.
Remove /-nsp/ for email.
On Sun, 19 Jun 2005 21:20:26 -0700, Paul Pluzhnikov wrote: Thomas Carter <T.******@nanob ots.com> writes:
void f(UInt32 * p)
{
UInt64 * q = (UInt64 *) p ;
*q = 1234ULL ;
return ;
}
with UInt64 typedef'd in the obvious way. Isn't there an unaligned access
on q now?
Yes, now there is. However, that code is now wrong, or at least
assumes that the caller passes a pointer to UInt64 without requiring
the caller to do so (which is a bad idea(TM)).
You claimed that such code is common (it isn't and it's obviously
broken) and that you've found such code in OpenSSL (did you?).
I admit to being carried away in my assessment here :-( My apologies.
Here is the case I have in mind:
A certain library uses char * throughout, with arbitrary alignments.
Since a char will be 1 byte long (at least, in general, I believe) such
pointers can have any alignment. Now if we have a low-level function that,
for performance reasons, and for a particular architecture, manipulates
unsigned int * as input, then we have a potential problem.
The code above, while obviously wrong, attempts to illustrate such case. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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