what I miss?

[Parahat Melayev]

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

( gcc version 4.0.0 20050519 (Red Hat 4.0.0-8) )

[Alexei A. Frounze]

"Parahat Melayev" <pa*****@gmail. com> wrote in message
news:11******** **************@ g47g2000cwa.goo glegroups.com.. .

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

Nope. C isn't a char, it's a pointer (to a char).

Alex

[Parahat Melayev]

oh yeah that is right :) panic

tnx

[ranjeet.gupta]

Parahat Melayev wrote:

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?
what is c ????
c is the pointer which is going to hold the address which is of the
char type.

malloc will return the address of the allocated memory.
and so c holds the address, which is int.

c = malloc(sizeof(u nsigned char));

HTH
ranjeet

( gcc version 4.0.0 20050519 (Red Hat 4.0.0-8) )

[Eric Sosman]

Parahat Melayev wrote:

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

Not very. I know of systems where this program
would report both sizes as zero -- a good deal stranger,
don't you think?

(Hint: What is the type of the result of `sizeof',
and what is the type expected by the "%d" conversion?)

--
Er*********@sun .com

[Parahat Melayev]

thanks but problem is not at malloc.
it must be,

printf("size of *c: %d\n", sizeof(*c));

[Kenny McCormack]

In article <11************ **********@f14g 2000cwb.googleg roups.com>,
Parahat Melayev <pa*****@gmail. com> wrote:

thanks but problem is not at malloc.
it must be,

printf("size of *c: %d\n", sizeof(*c));

Glad to see you've solved your own problem. Congratulations !

[Richard Heathfield]

Some minor nits...

Alexei A. Frounze wrote:

"Parahat Melayev" <pa*****@gmail. com> wrote in message
news:11******** **************@ g47g2000cwa.goo glegroups.com.. .

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

Nope. C isn't a char, it's a pointer (to a char).

No, C is a programming language (hint - case sensitivity!), and c is not a
pointer to a char, but a pointer to an unsigned char.

Usual stuff follows:

A) many people prefer a full prototype for main(), as in:
int main(void)

B) using the template p = malloc(n * sizeof *p), we could improve the malloc
to:
c = malloc(sizeof *c);
(ignoring n on this occasion, since we appear only to want one object).

C) sizeof yields a size_t, which is an unsigned integer type of unknown
size (in C90), so we will need to cast it. Unsigned longs are good
for this, so make that:

printf("size of unsigned char: %lu\n",
(unsigned long)sizeof(uns igned char));

(note that this is now required to write 1 on stdout), and

printf("size of c: %lu\n", (unsigned long)sizeof c);

Superfluous parentheses removed. Note the updated format specifiers.

Did I miss anything?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
mail: rjh at above domain

[pete]

ra***********@g mail.com wrote:

Parahat Melayev wrote:

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n",
sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this,
it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

what is c ????
c is the pointer which is going to hold the address which is of the
char type.

malloc will return the address of the allocated memory.
and so c holds the address, which is int.

The address is not int.
A pointer type is different from an int type.

--
pete

[Sensei]

On 2005-07-15 09:24:30 -0500, "Parahat Melayev" <pa*****@gmail. com> said:

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

int main()
{
unsigned char *c;
c = malloc(sizeof(u nsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
The size of an unsigned char is 1.

printf("size of c: %d\n", sizeof(c));
The size of a pointer is 4.

The size of the memory handled by the pointer is 1.

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn't that strange?

( gcc version 4.0.0 20050519 (Red Hat 4.0.0-8) )

I suggest reading a basic C book. You must know the difference between
a variable and a pointer.

PS. Sizes of vars and ptrs may vary on different platforms.

--
Sensei <se******@tin.i t>

cd /pub
more beer