Stupid cast question

[Dopey]

This is not homework.

I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
-- all I get is "$)\n". (I admit I'm an idiot, OK?)

#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
} ;

int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;

ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$' ;

for ( i = 0; i < sizeof( ms ) ; i++ )
buf[i] = (unsigned char)*((unsigne d char *)&ms[i]) ;

buf[i] = '\0' ;

printf( "%s\n", buf ) ;

return 0 ;
}

[Jens.Toerring]

Dopey <do***@nobrains .com> wrote:

This is not homework. I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
-- all I get is "$)\n". (I admit I'm an idiot, OK?) #include <stdio.h>
#include <stdlib.h> struct mystruct
{
int a, b, c, d, e, f, g ;
} ; int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;
Shouldn't you check that ms and buf aren't NULL?

ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$' ;

for ( i = 0; i < sizeof( ms ) ; i++ )
sizeof(ms) returns the size of the pointer, not the length of what
it points to. Use either "sizeof *ms" or sizeof(struct mystruct).
buf[i] = (unsigned char)*((unsigne d char *)&ms[i]) ;
ms isn't an array, but a pointer to a single structure (that's all
you allocated). Since you obviously try to print out the structure
in a byte by byte fashion you probably need

buf[i] = * ( ( unsigned char * ) msg + i );

to step over the memory pointed to by ms in a byte-by-byte way.

buf[i] = '\0' ;

printf( "%s\n", buf ) ;
This still won't print out "$$$$$$$\0\ n". First of all '\0' isn't a
printable character, it indicates the end of the string. Second,
the members of your mystruct are all ints, and since the size of an
int is usually larger than the one of a char you will have additional
(zero) bytes in between the '$'s (not even counting possible padding
bytes). The moment the printf() function finds one of the zero bytes
it will stop printing.
return 0 ;
}

Regards, Jens
--
_ _____ _____
| ||_ _||_ _| Je***********@p hysik.fu-berlin.de
_ | | | | | |
| |_| | | | | | http://www.physik.fu-berlin.de/~toerring
\___/ens|_|homs|_|oe rring

[Dan Pop]

In <MP************ ************@ne ws.verizon.net> Dopey <do***@nobrains .com> writes:

This is not homework.

I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
It was a good idea that you posted the expected output, because the
program is completely meaningless and we couldn't figure out how to fix
it without this piece of information.
-- all I get is "$)\n". (I admit I'm an idiot, OK?)
OK ;-)
#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
} ;

int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;
The size of the buffer is incorrectly calculated: you want one byte for
each member of the struct plus one byte for the string terminator.
Assuming a non-perverse implementation, you can obtain the number of
struct members (in this particular case) with the following expression:

sizeof *ms / sizeof ms -> a

or

sizeof(struct mystruct) / sizeof(int)

The former may look more cryptical, but it doesn't require any
maintenance if you decide to change the name of the struct or the type
of its members later.

So, the correct allocation for buf is:

unsigned char * buf = malloc((sizeof *ms / sizeof ms -> a) + 1 ) ;
ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$' ;

for ( i = 0; i < sizeof( ms ) ; i++ )
You got the iteration count horribly wrong: sizeof ms is giving you the
size of a pointer to a structure, which is completely unrelated to the
number of members of the struct. You need exactly the same expression
as above:

for (i = 0; i < sizeof *ms / sizeof ms -> a; i++)
buf[i] = (unsigned char)*((unsigne d char *)&ms[i]) ;
This is a horrible mess. You have allocated memory for a single
struct, not for a whole array of them. What you need is to convert ms
to pointer to int and use that pointer for accessing each member of
the struct, as if you had an array of int's:

buf[i] = ((int *)ms)[i] ;

As you can see, the correct expression is much simpler than your bit of
nonsense.
buf[i] = '\0' ;

printf( "%s\n", buf ) ;

return 0 ;
}

Now, let's see what we've got:

fangorn:~/tmp 2493> cat test.c
#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
};

int main(void)
{
int i;
struct mystruct *ms = malloc(sizeof(s truct mystruct));
char *buf = malloc((sizeof *ms / sizeof ms -> a) + 1);

ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$';
for (i = 0; i < sizeof *ms / sizeof ms -> a; i++)
buf[i] = ((int *)ms)[i];
buf[i] = '\0';
printf("%s\n", buf);
return 0 ;
}
fangorn:~/tmp 2494> gcc -ansi -pedantic test.c
fangorn:~/tmp 2495> ./a.out
$$$$$$$

It looks like the expected output. But the program still relies on the
reasonable assumption that struct mystruct contains no padding at all.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dopey]

Je***********@p hysik.fu-berlin.de wrote...

Dopey <do***@nobrains .com> wrote:

This is not homework.

I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
-- all I get is "$)\n". (I admit I'm an idiot, OK?)

#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
} ;

int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;

Shouldn't you check that ms and buf aren't NULL?

There's quite enough to criticize in the code without dinging me for
leaving the error checking out of the example I posted to illustrate
the problem.

[goose]

Dopey <do***@nobrains .com> wrote in message news:<MP******* *************** **@news.verizon .net>...

This is not homework.
even if it was, giving it a try before asking for help
is the accepted way of doing things.

<grin> its a good thing you follow accepted ways !

I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
-- all I get is "$)\n". (I admit I'm an idiot, OK?)
great, that'll save Dan Pop the effort of admitting that you
are an idiot :-)

#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
} ;

int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ; 97**@news.veriz on.net>...
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;
at this point, it is generally a good idea to check
that malloc returned a non-null pointer. what if the
system was out of memory ?
ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$' ;
ms->a = ms->b = ms->c = ms->d = ms->e = ms->f = ms->g = '$' ;
for ( i = 0; i < sizeof( ms ) ; i++ )
sizeof ( ms ) would be the size of a *pointer* to your struct.
to get the size of the actual object, you should use
sizeof (*ms);
buf[i] = (unsigned char)*((unsigne d char *)&ms[i]) ;
ms is *not* an array. it is a pointer to a single struct. I understand
what you are aiming for (i must, after all you posted the expected
output above :-), but I would recommend that you access each member
of a struct individually. the reason is because the compiler *may*
pad the struct, and accessing the struct as one block of memory is
bound to end in tears sooner or later, as you find 'gaps' in your
block of data would be filled with garbage.
uld be filled with garbage.
buf[i] = '\0' ;
'\0' ;
printf( "%s\n", buf ) ;
even if your compiler did not pad the structure, this can still
fail if sizeof (int) != sizeof (char), as printf ("%s\n", buf)
needs an array of chars, not an array of integers.

return 0 ;
}

hth

goose,
outlaw the non sc apps. makes life easier for *evryone* :-)

[pete]

Dopey wrote:

Je***********@p hysik.fu-berlin.de wrote...

Dopey <do***@nobrains .com> wrote:

This is not homework.

I'm trying to understand why this code doesn't output "$$$$$$$\0\ n"
-- all I get is "$)\n". (I admit I'm an idiot, OK?)

#include <stdio.h>
#include <stdlib.h>

struct mystruct
{
int a, b, c, d, e, f, g ;
} ;

int
main(void)
{
int i ;
struct mystruct * ms = malloc( sizeof(struct mystruct) ) ;
unsigned char * buf = malloc( sizeof(struct mystruct) + 1 ) ;

Shouldn't you check that ms and buf aren't NULL?

There's quite enough to criticize in the code without dinging me for
leaving the error checking out of the example I posted to illustrate
the problem.

It would have been less effort on your part
to include error checking originally, than to answer.
That you are personally aware that error checking belongs in the code,
is good, but it is not all that matters.
Your code is out there for all to see,
and it is missing error checking.

--
pete