sizeof dataTypes at run time

[Raman]

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Thanks and Regards,
Raman

[Peter Nilsson]

"Raman" <ramanchalo...@ gmail.comwrote:

Is ut possible to calculate size of any standard data types at
run time i.e without using sizeof() operator

http://groups.google.com/group/comp....rch+this+group

--
Peter

[Richard Heathfield]

Raman said:

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Sure - and this is especially easy since there is no such thing as the
sizeof() operator. All you have to do is instantiate the type, and use
the sizeof operator on the resulting object. This way, you avoid even
the appearance of using what someone has mistakenly described as a
sizeof() operator.

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

[jamsheedm]

On Mar 26, 10:54 am, "Raman" <ramanchalo...@ gmail.comwrote:

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Thanks and Regards,
Raman

Hi All,

U can get a sizeof any variable without even knowing what is its type
or where it is used. The way is explained below as an example.

int a;
printf("size of variable is %d\n",(&a+1)-(&a));

This is applicable for all datatypes(even for array).

Regards
jamsheed m(ja*******@gma il.com)

[Richard Heathfield]

ja*******@gmail .com said:

U can get a sizeof any variable without even knowing what is its type
or where it is used. The way is explained below as an example.

int a;
printf("size of variable is %d\n",(&a+1)-(&a));

This is applicable for all datatypes(even for array).

Wrong wrong wrong wrong wrong, as you'd know if you'd tried the above
code yourself.

The correct answer, always, is: use sizeof. That is what it is *for*.

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

[Richard Tobin]

In article <11************ **********@l75g 2000hse.googleg roups.com>,
ja*******@gmail .com <ja*******@gmai l.comwrote:

>int a;
printf("size of variable is %d\n",(&a+1)-(&a));

This is applicable for all datatypes(even for array).

.... and always prints 1.

-- Richard
--
"Considerat ion shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[ramana]

Yes it's possible to know. use this macro

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

where 'x' is the variable.

sample.c:
............... .......
#include <stdio.h>

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

int main()
{
char c;
int i;
float f;
double d;

printf(" sizes of char = %d \t int = %d \t float = %d \t double = %d
\n", SIZE(c), SIZE(i), SIZE(f), SIZE(d));

return 0;
}
............... ...............
Regards,
Ram

On Mar 26, 4:43 pm, "jamshe...@gmai l.com" <jamshe...@gmai l.comwrote:

On Mar 26, 10:54 am, "Raman" <ramanchalo...@ gmail.comwrote:

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Thanks and Regards,
Raman

Hi All,

U can get a sizeof any variable without even knowing what is its type
or where it is used. The way is explained below as an example.

int a;
printf("size of variable is %d\n",(&a+1)-(&a));

This is applicable for all datatypes(even for array).

Regards
jamsheed m(jamshe...@gma il.com)

[ramana]

yes. Its possible. Use the below macro.

sample program:
............... ............... .....
#include <stdio.h>

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

int main()
{
char c;
int i;
float f;
double d;

printf(" sizes of char = %d \t int = %d \t float = %d \t double = %d
\n", SIZE(c), SIZE(i), SIZE(f), SIZE(d));

return 0;
}
............... ............... ............

Regards,
Ram
Raman wrote:

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Thanks and Regards,
Raman

[Richard Heathfield]

ramana said:

yes. Its possible. Use the below macro.

sample program:
............... ............... ....
#include <stdio.h>

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

That doesn't calculate the size of a type.

Proof:

#include <stdio.h>

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

int main(void)
{
printf("%d\n", (int)SIZE(doubl e));
return 0;
}

foo.c: In function `main':
foo.c:7: parse error before `double'
foo.c:7: parse error before `)'

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

[Jack Klein]

On 26 Mar 2007 06:14:28 -0700, "ramana" <ra************ *@gmail.com>
wrote in comp.lang.c:

Do not top post in this group, it is considered rude. Text you add
should be after or interspersed with quoted text you are responding
to. I have reformatted your post and added my comments.

On Mar 26, 4:43 pm, "jamshe...@gmai l.com" <jamshe...@gmai l.comwrote:

On Mar 26, 10:54 am, "Raman" <ramanchalo...@ gmail.comwrote:

Hi All,

Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

Thanks and Regards,
Raman

Hi All,

U can get a sizeof any variable without even knowing what is its type
or where it is used. The way is explained below as an example.

int a;
printf("size of variable is %d\n",(&a+1)-(&a));

This is applicable for all datatypes(even for array).

Regards
jamsheed m(jamshe...@gma il.com)

Yes it's possible to know. use this macro

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

where 'x' is the variable.

sample.c:
............... ......
#include <stdio.h>

#define SIZE(x) ( ((char*)(&x + 1)) - (char*)&(x) )

The macro is correct.

int main()
{
char c;
int i;
float f;
double d;

printf(" sizes of char = %d \t int = %d \t float = %d \t double = %d
\n", SIZE(c), SIZE(i), SIZE(f), SIZE(d));

The printf() statement above is non-portable at best, and causes
undefined behavior at worst. The subtraction of two pointers yields a
result with a type of ptrdiff_t, which is a signed integer type but
not necessarily a signed int.

If ptrdiff_t happens to be signed int on your implementation, or
signed short (unlikely!), then the code will do what you want.

On implementations where ptrdiff_t is a wider type than int, and I
work with some, the code produces undefined behavior.

The obvious solution is to add a cast to int in your macro. Less
obvious but slightly better is to add a cast to long, and change the
conversion specifiers to %ld.

Best, of course, is get rid of the macro and use sizeof.

return 0;
}
............... ..............

Kindly learn how to post a proper signature line. It is separated
from the body of your message by the character string "-- " appearing
on a line by themselves. See my signature below for an example.

Regards,
Ram

--
Jack Klein
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