sizeof struct array

[ak]

struct xy{
int x;
int y;
}
_xy[32];

size_of_xy(stru ct xy * a) {
int len = sizeof a;
printf( "\sizeof xy: %i\n", len );
}

int main( int argc, char ** argv ) {
size_of_xy(_xy) ;
}

the problem is that size_of_xy prints size of xy struct.
I need to print size of _xy array.

any ideas?

Andrei

[S.Tobias]

ak <sp**@imagero.c om> wrote:

struct xy{
int x;
int y;
}
_xy[32]; size_of_xy(stru ct xy * a) {
int len = sizeof a;
ITYM sizeof *a;
The result of `sizeof' is type `size_t'.
printf( "\sizeof xy: %i\n", len );
} int main( int argc, char ** argv ) {
size_of_xy(_xy) ;
} the problem is that size_of_xy prints size of xy struct.
I need to print size of _xy array.
No way.
any ideas?

You have to pass the length of the array in another argument.

If you want more data encapsulation, you might think of sth like:

struct xy_array {
struct xy *xy_arr;
size_t xy_arr_len;
}; /*you are responsible for its contents*/

size_of_xy_arra y(struct xy_array *a) {
printf("size: %u\n", (unsigned)(size of(struct xy) * a->xy_arr_len)) ;
}

--
Stan Tobias
sed '/[A-Z]//g' to email

[Martin Ambuhl]

ak wrote:

struct xy{
int x;
int y;
}
_xy[32];

size_of_xy(stru ct xy * a) {
int len = sizeof a;
printf( "\sizeof xy: %i\n", len );
}

int main( int argc, char ** argv ) {
size_of_xy(_xy) ;
}

the problem is that size_of_xy prints size of xy struct.
I need to print size of _xy array.

any ideas?

/* mha: you could read the FAQ before posting. Or if that's beyond
you, check the archives at groups.google.c om. Those are the things
civilized people do before posting. In any case, consider the code
below */
#include <stdio.h>

#define array_size_info (a) do \
printf("The array has size %lu,"\
" each of the %lu elements has size %lu\n",\
sizeof a, sizeof a/sizeof *a, sizeof *a);\
while (0)

int main(void)
{
struct xy
{
int x;
int y;
}
xy[32];
array_size_info (xy);
return 0;
}
[output on my implementation]

The array has size 256, each of the 32 elements has size 8

[ak]

> /* mha: you could read the FAQ before posting. Or if that's beyond

you, check the archives at groups.google.c om. Those are the things
civilized people do before posting. In any case, consider the code
below */
#include <stdio.h>

#define array_size_info (a) do \
printf("The array has size %lu,"\
" each of the %lu elements has size %lu\n",\
sizeof a, sizeof a/sizeof *a, sizeof *a);\
while (0)

int main(void)
{
struct xy
{
int x;
int y;
}
xy[32];
array_size_info (xy);
return 0;
}
[output on my implementation]

The array has size 256, each of the 32 elements has size 8

#include <stdio.h>

#define array_size_info (a) do \
printf("The array has size %lu,"\
" each of the %lu elements has size %lu\n",\
sizeof a, sizeof a/sizeof *a, sizeof *a);\
while(0)

struct xy {
int x;
int y;
}
first_xy[32], second_xy[64];

void do_something(st ruct xy *_xy) {
array_size_info (_xy);
}

int main( void ) {
array_size_info ( first_xy );
array_size_info ( second_xy );
do_something(fi rst_xy);
do_something(se cond_xy);
return 0;
}

the output is:

The array has size 256, each of the 32 elements has size 8
The array has size 512, each of the 64 elements has size 8
The array has size 4, each of the 0 elements has size 8
The array has size 4, each of the 0 elements has size 8

as you can see array_size_info works wrong in do_something();

--
Andrei Kouznetsov

[Martin Ambuhl]

ak wrote:

/* mha: you could read the FAQ before posting. Or if that's beyond
you, check the archives at groups.google.c om. Those are the things
civilized people do before posting. In any case, consider the code
below */
#include <stdio.h>

#define array_size_info (a) do \
printf("The array has size %lu,"\
" each of the %lu elements has size %lu\n",\
sizeof a, sizeof a/sizeof *a, sizeof *a);\
while (0)

int main(void)
{
struct xy
{
int x;
int y;
}
xy[32];
array_size_info (xy);
return 0;
}
[output on my implementation]

The array has size 256, each of the 32 elements has size 8

#include <stdio.h>

#define array_size_info (a) do \
printf("The array has size %lu,"\
" each of the %lu elements has size %lu\n",\
sizeof a, sizeof a/sizeof *a, sizeof *a);\
while(0)

struct xy {
int x;
int y;
}
first_xy[32], second_xy[64];

void do_something(st ruct xy *_xy) {
array_size_info (_xy);
}

int main( void ) {
array_size_info ( first_xy );
array_size_info ( second_xy );
do_something(fi rst_xy);
do_something(se cond_xy);
return 0;
}

the output is:

The array has size 256, each of the 32 elements has size 8
The array has size 512, each of the 64 elements has size 8
The array has size 4, each of the 0 elements has size 8
The array has size 4, each of the 0 elements has size 8

as you can see array_size_info works wrong in do_something();

It works fine. You gave it a broken argument. I suggested that you
should have read the FAQ. I repeat that. If you can't figure out that
the argument given to array_size_info in do_something is a pointer and
not an array, may God have mercy on your soul.

[ak]

> If you can't figure out that

the argument given to array_size_info in do_something is a pointer and
not an array, may God have mercy on your soul.

sure, he has.

why I can't get size of array if I have only pointer to it?

--
Andrei

[Alex Fraser]

"ak" <sp**@imagero.c om> wrote in message news:cd******** **@online.de...

If you can't figure out that
the argument given to array_size_info in do_something is a pointer and
not an array, may God have mercy on your soul.

sure, he has.

why I can't get size of array if I have only pointer to it?

Because a pointer to an array holds a value that tells you where the start
of the array is, and nothing else (such as the size).

This is why, if you need it, you must pass the size of the array in another
argument. You might not always need the size; sometimes you can use a
sentinel value to indicate the end. A good example of this is strings in the
standard library, where '\0' is used as the sentinel value.

Alex

[Martin Ambuhl]

ak wrote:

If you can't figure out that
the argument given to array_size_info in do_something is a pointer and
not an array, may God have mercy on your soul.

sure, he has.

why I can't get size of array if I have only pointer to it?

I suggestes in my first reply, and repeated in the second, that you
should look to the FAQ, which you have obviously not done. How big a
hint do you need?
If you would *READ THE DAMN FAQ* you wouldn't keep asking these stupid
questions.

[ak]

> I suggestes in my first reply, and repeated in the second, that you

should look to the FAQ, which you have obviously not done. How big a
hint do you need?
If you would *READ THE DAMN FAQ* you wouldn't keep asking these stupid
questions.

Stay cool.
I'll make it. Sometimes.

--
Andrei

[Al Bowers]

ak wrote:

I suggestes in my first reply, and repeated in the second, that you
should look to the FAQ, which you have obviously not done. How big a
hint do you need?
If you would *READ THE DAMN FAQ* you wouldn't keep asking these stupid
questions.

Stay cool.
I'll make it. Sometimes.

That's right! Keep cool.

You could make this array a typedef and then when your function
carries a pointer to this typedef, you can dereference it
for the size.

#include <stdio.h>

typedef struct _XY
{
int x;
int y;
} _XY[32];

void size_of_xy(_XY *a)
{
printf("The array has size %u,"
" Each of the %u elements has size %u\n"
"The pointer to the array has size %u\n",
sizeof *a,sizeof *a/sizeof(struct _XY),
sizeof(struct _XY),sizeof a);
}

int main(void)
{
_XY _xy;
size_of_xy(&_xy );
return 0;
}

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/