pointer to string and then back to pointer on a 64 bit machine.

[CBFalconer]

Re: pointer to string and then back to pointer on a 64 bit machine.

Tor Rustad wrote:

CBFalconer wrote:
>
[...]
>

>No. All pointers can be cast to void* and back to the same type.

>
Nope. All pointers to objects can, but there is no such guarantee
for function pointers.

True. I know that, but never seem to remember to mention it.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>



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[Barry Schwarz]

Re: pointer to string and then back to pointer on a 64 bit machine.

On Thu, 27 Sep 2007 15:32:05 -0000, let_the_best_ma n_win
<sgiitnewid@gma il.comwrote:

>On Sep 27, 7:42 pm, CBFalconer <cbfalco...@yah oo.comwrote:

>let_the_best_m an_win wrote:
>>

How do I print a pointer address into a string buffer and then
read it back on a 64 bit machine ?

>>
>Use %p. Make sure the pointer is cast to void* for printing.

>
>Why this restriction of casting to a void* ?

Because the answer to the next question is no.

>Shouldn't the size of all pointers be same ?

The only requirement is that void* and char* have the same size,
alignment, and representation. Other types of pointers are not
constrained this way.

>(Now please do not go on to HUGE and LONG pointers, whatever they
>are !)

There are no HUGE or LONG pointers in C. There are pointers to long
but C is case sensitive. And a long* need not be the same size as any
other pointer.


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[Charlie Gordon]

Re: pointer to string and then back to pointer on a 64 bit machine.

"Martin Ambuhl" <mambuhl@earthl ink.neta écrit dans le message de news:
5m1ipfFb4a26U1@ mid.individual. net...

let_the_best_ma n_win wrote:

>How do I print a pointer address into a string buffer and then read it
>back on a 64 bit machine ?
>Compulsion is to use string (char * buffer) only.
>>
>printing with %d does not capture the full 64-bits of the pointer.
>does %l exist in both printf and scanf for this purpose ?

>
Your references to a "64 bit machine" suggest that you don't really
understand pointers. They are to an address space for which the values
may not correspond to any physical address on your machine.
>
And "%d" is a specifier for signed ints, not for pointers. Check the code
below for hints about how to do what you _may_ (or may not) be meaning to
do. Your question is underspecified, as I'm sure you will realize on
reflection.
>
#include <stdio.h>
#include <stdlib.h>
>
int main(void)
{
char *buffer;
size_t n;
>
/* find out what the length of the string needs to be */
n = snprintf(0, 0, "%p", (void *) &buffer);

You compute the length of the representation of &buffer, which is the
address of the pointer, not the pointer itself.

/* allocate the space */
if (!(buffer = malloc(n + 1))) {
fprintf(stderr, "could not allocate space for buffer.\n"
"giving up ...\n");
exit(EXIT_FAILU RE);
}

good, now buffer has a value, but there is no guarantee it is large enough
for the representation of its own address by %p.

/* put the address of buffer into buffer and show it */
sprintf(buffer, "%p", (void *) buffer);

bingo! potential buffer overflow

printf("The buffer is at %p, and contains the string \"%s\"\n",
(void *) buffer, buffer);
>
free(buffer);
return 0;
}
>
The buffer is at 20d98, and contains the string "20d98"

The above code is broken.

--
Chqrlie.