struggling to use calloc and realloc

[dagger]

Hi there. I'm using C under FreeBSD with the gcc compiler and am having a
bit of trouble using the calloc and realloc calls.

As an example the code snippet:

#include <stdio.h>

int main() {

char *ptr;

ptr = (char *) calloc(1, sizeof(char));

printf( "initial size (1 char) = %d\n", sizeof(ptr) );

ptr = (char *) realloc(ptr, sizeof(char)*10 );

printf( "new size (10 chars) = %d\n", sizeof(ptr) );

return 0;
}
and yet when I run it I get a size of 4 for each printf even though
initially I allocated only the size of 1 char, and after reallocation there
should be room for 10 bytes...? Or where am I making a mistake in my
reasoning?

Thanks for any help :)

[Dave Vandervies]

In article <3f********@new s1.mweb.co.za>, dagger <fe****@mweb.co .za> wrote:

Hi there. I'm using C under FreeBSD with the gcc compiler and am having a
bit of trouble using the calloc and realloc calls.

As an example the code snippet:
[trimmed mercilessly] char *ptr;
printf( "initial size (1 char) = %d\n", sizeof(ptr) );
printf( "new size (10 chars) = %d\n", sizeof(ptr) ); and yet when I run it I get a size of 4 for each printf even though
initially I allocated only the size of 1 char, and after reallocation there
should be room for 10 bytes...? Or where am I making a mistake in my
reasoning?

sizeof(ptr) gives you the size of the pointer (and getting 4 for that
means you're running the program on a machine where pointers are 4 bytes
wide, which in this day and age suggests a desktop machine running a
32-bit operating system).
There's no way to get back the size of the memory a pointer points to;
"sizeof *ptr" will give the size of one of whatever type ptr points at,
but (especially for memory obtained from malloc and its friends) your
code needs to keep track of how many of those it has space for itself.
dave

--
Dave Vandervies dj******@csclub .uwaterloo.ca
Yes, I suppose it could if it really wanted. I stand corrected. Of course,
I'd question such a compiler's morals and drinking habits...
--Richard Heathfield in comp.lang.c

[Servé Lau]

"dagger" <fe****@mweb.co .za> wrote in message
news:3f******** @news1.mweb.co. za...

Hi there. I'm using C under FreeBSD with the gcc compiler and am having a
bit of trouble using the calloc and realloc calls.

As an example the code snippet:

#include <stdio.h>
Didn't you get a warning? For realloc and calloc you need stdlib.h too.
int main() {

char *ptr;

ptr = (char *) calloc(1, sizeof(char));
Ah, you didn't get a warning because of the cast, don't cast the return
value of the alloc functions.
printf( "initial size (1 char) = %d\n", sizeof(ptr) );
sizeof(ptr) is the same as sizeof(char *), on your system sizeof(char *) is
4 bytes.
and yet when I run it I get a size of 4 for each printf even though
initially I allocated only the size of 1 char, and after reallocation there should be room for 10 bytes...? Or where am I making a mistake in my
reasoning?

sizeof gives the size of the variable itself, not the size of its value.

You wouldn't expect this either.

int x = 100;
if (sizeof x == 100) false ot true?

[Christopher Benson-Manica]

dagger <fe****@mweb.co .za> spoke thus:

#include <stdio.h>
int main() {
char *ptr;
ptr = (char *) calloc(1, sizeof(char)); ^^^^^^^^
The cast isn't needed, and will hide helpful compiler warnings should
you have the misfortune to forget to include stdio.h. You should
check that the allocation succeeded before continuing. Why not just
use malloc()? And you might want to do

malloc( sizeof(*ptr) );
printf( "initial size (1 char) = %d\n", sizeof(ptr) );
ptr is a pointer to a character. sizeof(ptr) is the space required
by such an entity; on your implementation, it happens to be four
bytes. This number has no relation to how many bytes of allocated
memory, if any, ptr points to.
ptr = (char *) realloc(ptr, sizeof(char)*10 );
Again, lose the cast. If the original allocation succeeds, but this
one fails, the memory you originally allocated is now gone forever.
Make sure you can free() what you allocated in this event:

char *ptr, *tmp;

ptr=malloc( sizeof(*ptr) );
tmp=ptr;
ptr=realloc( ptr, 10*sizeof(*ptr) );
if( !ptr ) {
free( tmp ); /* tmp pointed at the memory originally malloc()'ed,
and that memory is now freed */
/* go on with life */
}
else {
/* do what you want with ptr */
}
printf( "new size (10 chars) = %d\n", sizeof(ptr) );

Same error as above.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Christopher Benson-Manica]

Christopher Benson-Manica <at***@nospam.c yberspace.org> spoke thus:

ptr = (char *) calloc(1, sizeof(char));

The cast isn't needed, and will hide helpful compiler warnings should
you have the misfortune to forget to include stdio.h. You should

^^^^^^^

Of course, what would a post from me be like without some silly error
or other? This is, of course, meant to be stdlib.h. I need a post
preprocessor to catch these nasssties...

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[nrk]

dagger wrote:

Hi there. I'm using C under FreeBSD with the gcc compiler and am having a
bit of trouble using the calloc and realloc calls.

As an example the code snippet:

#include <stdio.h>

int main() {

char *ptr;

ptr = (char *) calloc(1, sizeof(char));

printf( "initial size (1 char) = %d\n", sizeof(ptr) );

ptr = (char *) realloc(ptr, sizeof(char)*10 );

printf( "new size (10 chars) = %d\n", sizeof(ptr) );

return 0;
}
and yet when I run it I get a size of 4 for each printf even though
initially I allocated only the size of 1 char, and after reallocation
there
should be room for 10 bytes...? Or where am I making a mistake in my
reasoning?

Thanks for any help :)

sizeof operator determines the size of the type of its operand. In this
case the type of the operand (ptr) is char *, and the size of a char * on
your machine happens to be 4, which is what's being printed. sizeof will
not tell you how many valid bytes ptr is pointing to (that is, it won't
tell you what was allocated using calloc/malloc/realloc). This information
cannot be extracted from the pointer. You need to keep track of the
allocated size in a separate variable.

-nrk.

[Kevin Goodsell]

dagger wrote:

Hi there. I'm using C under FreeBSD with the gcc compiler and am having a
bit of trouble using the calloc and realloc calls.

As an example the code snippet:

#include <stdio.h>

int main() {

char *ptr;

ptr = (char *) calloc(1, sizeof(char));
A few notes here: first, sizeof(char) is a fancy way of writing 1.
Second, calloc() is not a useful as it seems. Strictly speaking, it may
not give the objects you allocate the value you expect. This is
particularly true for floating point and pointer types. I need to check
the standard to be sure, but I've been told that all-bits-zero is not
guaranteed to be the representation of 0 for any type other than the
character types.

Finally, casting the return from malloc & friends is not recommended,
and your example seems to demonstrate one of the reasons why. You forgot
to #include <stdlib.h>, which means that there is no prototype in scope
for calloc. This in turn means that the compiler assumed (incorrectly) a
return type of int. ints and pointers may not be returned the same way,
so the pointer that calloc returned might be lost, and ptr might be
assigned garbage. Without the cast, the compiler would have been
required to warn you about the conversion from int to a pointer. With
the cast, there is no such guarantee.

Consider using the comp.lang.c-approved idiom for calling malloc & friends:

ptr = calloc(1, sizeof *ptr);

This has number of advantages. The most important are that it doesn't
mask the error mentioned above, it's largely self-maintaining if the
type of ptr changes (for example, of you later choose to support
internationaliz ation using the type wchar_t instead of char), and it's
more difficult to screw up than other idioms.

printf( "initial size (1 char) = %d\n", sizeof(ptr) );
There are at least 2 apparent problems here. First, you seem to be
expecting sizeof to tell you how many characters you've allocated. It
cannot do that. sizeof is purely a compile-time construct. Here it will
give the size of the pointer ptr. If you want the size of a dynamically
allocated object, you need to keep track of it yourself.

The other problem is that you are using the wrong format specifier. %d
can never be an appropriate format specifier for type size_t. %d is ONLY
for type (signed) int. size_t cannot be an alias for (signed) int
because size_t must be unsigned.

Keep in mind that printf() is stupid. It can't determine the types of
the objects you pass to it. It relies completely on you telling it what
the types are. If you lie, all bets are off. Stack corruption leading to
a program crash or worse is a likely result. Always double-check your
format strings.

For size_t, you can cast to unsigned long and print using %lu. This is
mostly safe, but could potentially give incorrect results in C99, if the
size_t value exceeds the range of unsigned long.

ptr = (char *) realloc(ptr, sizeof(char)*10 );
This is not a safe way of using realloc. If it fails and returns NULL,
you leak whatever memory you had already allocated. To realloc safely,
store the result into a temporary pointer, then assign into your
original pointer only if the result is non-NULL.

Also, all the notes about calloc apply here as well (except that realloc
doesn't initialize the bytes to all-bits-0).

printf( "new size (10 chars) = %d\n", sizeof(ptr) );
Same problems as the previous printf().

return 0;
}
and yet when I run it I get a size of 4 for each printf even though
initially I allocated only the size of 1 char, and after reallocation there
should be room for 10 bytes...? Or where am I making a mistake in my
reasoning?

Yes. You are printing the sizeof a pointer, not a dynamically allocated
object. C provides no way of determining the size of a dynamically
allocated object, post-allocation.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Christopher Benson-Manica]

Kevin Goodsell <us************ *********@never box.com> spoke thus:

all-bits-zero is not
guaranteed to be the representation of 0 for any type other than the
character types.

It may be a trap representation for integer types, yes?

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Kevin Goodsell]

Christopher Benson-Manica wrote:

Kevin Goodsell <us************ *********@never box.com> spoke thus:

all-bits-zero is not
guaranteed to be the representation of 0 for any type other than the
character types.

It may be a trap representation for integer types, yes?

I think I've heard conflicting reports from different experts, or
possibly I've misunderstood some of what I've heard. It's one of those
things I've been meaning to look up for myself, but I'm afraid it'll
take me hours to find and decrypt all the relevant sections. :/

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Alex]

Christopher Benson-Manica <at***@nospam.c yberspace.org> wrote:

Kevin Goodsell <us************ *********@never box.com> spoke thus:

all-bits-zero is not
guaranteed to be the representation of 0 for any type other than the
character types.

It may be a trap representation for integer types, yes?

Signed integer types. Yes.

Alex