obtaining the size of a structure

[kris]

Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );
}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

Thanks
Krish

[Mark Bluemel]

kris wrote:

Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

Standard headers such as <stdio.hmissi ng

struct abc
{
int j;
char k;
int i;
}*a;

main()

I'd prefer "int main(void)" and I think it may be required in C99

{
a->j=10;

using the value of a uninitialized pointer. Bad dog! No biscuit!

>
printf("size of structure is %d",sizeof(a) );

That tells you the size of the pointer. "sizeof(*a) " would tell you the
size of the structure.

}

No "return" statement is poor style.

[Nikos Chantziaras]

kris wrote:

Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );
}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

(That program won't work, but I assume it's just a copy&paste error?)

First, sizeof(a) will always return the number of bytes occupied by
pointers (since "a" is a pointer), which should be 4 on most 32-bit systems.

Second, the size of the struct is "sizeof(str uct abc)". This is the
same as "sizeof(*a) ", assuming you allocated the space for "a" with:

a = malloc(sizeof(s truct abc));

The size of the struct is always the same. I don't understand what you
mean with "I need to get the size of structure such that it resembles
the chuck of memory that is actually being used".

[Mark Bluemel]

Nikos Chantziaras wrote:

kris wrote:

>Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );
}

Second, the size of the struct is "sizeof(str uct abc)". This is the
same as "sizeof(*a) ", assuming you allocated the space for "a" with:

a = malloc(sizeof(s truct abc));

Nope - sizeof(*a) will return the sizeof the structure, even for an
uninitialized or NULL pointer...

[Fred Kleinschmidt]

"kris" <ra************ **@gmail.comwro te in message
news:11******** *************@i 38g2000prf.goog legroups.com...

>
Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );

This is the size of a pointer to the structure.
'a' is of type 'struct abc *'

}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

try sizeof(*a)

Why do you need this?

Note that the actual size of this
structure is implementation dependent; the compiler may
add padding between j and k, and between k and i, and
even after i. Different compilers may add different
amounts of padding.
--
Fred L. Kleinschmidt

[santosh]

kris wrote:

>
Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

a is a pointer of type struct abc*.

main()

As per the C Standard main must return an int. So declare it to do so.

int main(void)

{
a->j=10;

a is a pointer. You need to initialise it to point to a valid structure abc
object. Do;

struct abc foo;
a = &foo;
a->j = 10;

printf("size of structure is %d",sizeof(a) );

The d format specifier is for an int argument. Use either lu or zu for a
size-t argument, which is the return type of the sizeof operator.

To calculate the size of the structure do:
sizeof *a;

Note the lack of parenthesis.

}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

No you don't. You get the size of the pointer a.

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

Do as above.

[Army1987]

On Wed, 01 Aug 2007 07:09:35 -0700, kris wrote:

>
Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()

In C99, you need to specify the return type (int).

{
a->j=10;

printf("size of structure is %d",sizeof(a) );

sizeof returns a size_t, which cannot be an int (as expected by
%d). See www.c-faq.com, question 12.9b.

In C89, you need to explicitly return a value (0 to indicate
success).

}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

No, you get the size of a pointer to it.

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

sizeof *a, i.e. sizeof(struct abc) is required to include the
padding.
(a is a pointer to struct abc, so *a is a struct abc.)
--
Army1987 (Replace "NOSPAM" with "email")
"Never attribute to malice that which can be adequately explained
by stupidity." -- R. J. Hanlon (?)

[Army1987]

On Wed, 01 Aug 2007 17:35:38 +0300, Nikos Chantziaras wrote:

kris wrote:

>Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );
}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

(That program won't work, but I assume it's just a copy&paste error?)

First, sizeof(a) will always return the number of bytes occupied by
pointers (since "a" is a pointer), which should be 4 on most 32-bit systems.

Second, the size of the struct is "sizeof(str uct abc)". This is the
same as "sizeof(*a) ", assuming you allocated the space for "a" with:

a = malloc(sizeof(s truct abc));

The size of the struct is always the same. I don't understand what you
mean with "I need to get the size of structure such that it resembles
the chuck of memory that is actually being used".

I think he means the size of the struct including the padding,
rather than just the sum of the sizes of its members.
--
Army1987 (Replace "NOSPAM" with "email")
"Never attribute to malice that which can be adequately explained
by stupidity." -- R. J. Hanlon (?)

[santosh]

kris wrote:

>
Hi I am writing a small program where I need to obtain the actual size
of a structure.
The programm is as follows

struct abc
{
int j;
char k;
int i;
}*a;

main()
{
a->j=10;

printf("size of structure is %d",sizeof(a) );
}

In this program when I print the size of structure I get the whole
size (i.e size of 2 int variables+size of char variable).

But I need to get the size of structure such that it resembles the
chunk of memory that is actually being used.

Don't post multiple times. Usenet, to which Google Groups interfaces, can
sometimes be slow. Wait for a few hours before reposting. See you earlier
post for replies.

[JT]

Army1987 wrote:

Also, a compiler
isn't required to let you specify the alignment.

On Aug 1, 3:25 pm, jacob navia <[e-mail] mcomp.frwrote:

Ahh that is news to me. How would you do it then?
(just curious how you do it without a compiler)

I'm sure he meant
"a compiler doesn't have to let you specify the alignment", i.e.
"a compiler isn't required to let you specify the alignment".

And he would be right.