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

[Ben Pfaff]

ilan pillemer <il**@pillemer. netwrites:

But this is what I did. I set the bit values of the basic types to ~0.
For unsigned basic types this gave me the range.

It won't necessarily work for unsigned integer types wider than
int. For these you'll need to either use the proper type for 0,
as in e.g. ~0UL. Or you can just assign -1 to an object of the
unsigned integer type, which doesn't require such additional
care.

By the way, it's best to think of C objects as having values, not
representations or "bit values". Most operations in C operate on
values, not on representations .

And signed the maximum negative value. I could also determine
thus by a comparison if 2's complement or 1's complement was
being used. I thus had my answer through direct computation.

I think that ~0 can actually yield a trap representation.
--
Comp-sci PhD expected before end of 2007
Seeking industrial or academic position *outside California* in 2008

[Keith Thompson]

ilan pillemer <il**@pillemer. netwrites:

"Army1987" <pl********@for .itwrites:

>"Raman" <ra***********@ gmail.comha scritto nel messaggio
news:11******* *************** @l75g2000hse.go oglegroups.com. ..

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

I believe this is an excercise in K&R2,p 36.

"Exercise 2-1. Write a program to determine the ranges of char,
short, int and long variables, both signed and unsigned, by printing
appropriate values from standard headers and by direct
computation. Harder if you compute them: determine the ranges of
various floating point types."

[...]

The exercise computes the *ranges* of the types (e.g., 0..65535), not
their sizes (e.g., 2 bytes or 16 bits).

(These are examples; I'm not assuming that a byte is 8 bits.)

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[ilan pillemer]

Keith Thompson <ks***@mib.orgw rites:

ilan pillemer <il**@pillemer. netwrites:

"Army1987" <pl********@for .itwrites:

"Raman" <ra***********@ gmail.comha scritto nel messaggio
news:11******** **************@ l75g2000hse.goo glegroups.com.. .
Is ut possible to calculate size of any standard data types at run
time i.e without using sizeof() operator

I believe this is an excercise in K&R2,p 36.

"Exercise 2-1. Write a program to determine the ranges of char,
short, int and long variables, both signed and unsigned, by printing
appropriate values from standard headers and by direct
computation. Harder if you compute them: determine the ranges of
various floating point types."

[...]

The exercise computes the *ranges* of the types (e.g., 0..65535), not
their sizes (e.g., 2 bytes or 16 bits).

(These are examples; I'm not assuming that a byte is 8 bits.)

...hmm.. Would I look the at the number of bits in a char to determine
how many bits in a byte?

If this is a correct I could then fill a unsigned char with bits set
to one and then left shift them until all the bits are zeroed?

--
ilAn

[Richard Heathfield]

ilan pillemer said:

Keith Thompson <ks***@mib.orgw rites:

<snip>

>>
The exercise computes the *ranges* of the types (e.g., 0..65535), not
their sizes (e.g., 2 bytes or 16 bits).

(These are examples; I'm not assuming that a byte is 8 bits.)

..hmm.. Would I look the at the number of bits in a char to determine
how many bits in a byte?

Yes, because a char is exactly one byte wide. Incidentally, it is also
exactly CHAR_BIT bits wide.

If this is a correct I could then fill a unsigned char with bits set
to one and then left shift them until all the bits are zeroed?

Yes, you could do that, or you could simply do this:

unsigned char x = 1;
int char_bit = 0;
while(x 0)
{
++char_bit;
x <<= 1;
}

or of course you could just use CHAR_BIT from <limits.h>.

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

[ilan pillemer]

Richard Heathfield <rj*@see.sig.in validwrites:

ilan pillemer said:

Keith Thompson <ks***@mib.orgw rites:

<snip>

>
The exercise computes the *ranges* of the types (e.g., 0..65535), not
their sizes (e.g., 2 bytes or 16 bits).

(These are examples; I'm not assuming that a byte is 8 bits.)

..hmm.. Would I look the at the number of bits in a char to determine
how many bits in a byte?

Yes, because a char is exactly one byte wide. Incidentally, it is also
exactly CHAR_BIT bits wide.

If this is a correct I could then fill a unsigned char with bits set
to one and then left shift them until all the bits are zeroed?

Yes, you could do that, or you could simply do this:

unsigned char x = 1;
int char_bit = 0;
while(x 0)
{
++char_bit;
x <<= 1;
}

or of course you could just use CHAR_BIT from <limits.h>.

I know this. I am also learning from "Advanced Programmig in the UNIX
environment." On page 27, figure 2.6 Stevens/Rago lists 19 limits that
can be found in <limits.h>. The K&R2 exercise said I should also as an
exercise determine this kind of information by direct
computation. (Excerise 2-1 on page 36). That is why I am trying to do
this without referring to <limits.h>.

--
ilAn