How can I compute the range of signed and unsigned types

[Ahmed Masud]

Errrrr... :-) have fun


#include <stdio.h>

#define BYTESIZE(me)    (sizeof(me))
#define BITSIZE(me)     (2 << sizeof(me))

int main() {
        fprintf(stderr, "%d:%d\n", BYTESIZE(char), BITSIZE(char));
        fprintf(stderr, "%d:%d\n", BYTESIZE(short), BITSIZE(short));
        fprintf(stderr, "%d:%d\n", BYTESIZE(int), BITSIZE(int));
        fprintf(stderr, "%d:%d\n", BYTESIZE(long), BITSIZE(long));
}


On Wed, 18 Apr 2001, Jared Hodge wrote:

> Chris,
> 	The specification for C and C++ aren't specific about the size of data
> types, but I believe that for other languages they are (Fortran
> perhaps).  In C++ I've seen the data size of say integers vary from 16
> to 32 bits from one version of a compiler to the next.  They almost
> always use 2's complement for signed integers however, and it is very
> straightforward to compute the possible data size of a known bit width
> integer.  (In the following formulas I'll use ^ to represent a
> superscript):
> 
> For a 2s complement signed integer: The max value is the sum of two to
> the power of all of the lower order bit placements
> For an 8 bit integer:
> Max = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 2^8 - 1
> For a 16 bit integer: Max = 2^16 - 1
> etc.
> 
> What 2's complement means is that the highest order bit is negative 2 to
> that power, so the lowest number that is possible is
> For an 8 bit integer: Min = - 2^8
> For a 16 bit integer: Min = - 2^16
> etc.
> 
> For an unsigned integer the highest order bit is used for positive
> numbers also, so you include it in the total listed above for signed
> integers
> For an 8 bit unsigned integer: Max = 2^8 - 1 + 2^8 = 2^9 - 1
> For a 16 bit unsigned integer: Max = 2^17 - 1
> 
> The Min for all unsigned integers is zero (since the high order bit is
> used for positive numbers).
> 
> This problem is not near as simple for floating point integers, since
> the number of bits used is split between the prefix number and exponent
> (i.e. similar to scientific notation 3.134 * 10^-2, 3.134 is the prefix,
> -2 is the exponent).  This is why their is a bound on the accuracy of a
> floating point number more than the range (although there is definitely
> a range bound).
> 
> I hope this answers your question.  Let me know if you have any more
> questions.
> 
> 
> Chris Richard Adams wrote:
> > 
> > This is my point - how can I compute this so my code could run on any
> > machine. I need to show the range possible on any machine...how can I
> > compute that?
> > 
> > Thanks,
> > Chris
> > 
> > -----Original Message-----
> > From: James Cownie [mailto:jcownie at etnus.com]
> > Sent: Wednesday, April 18, 2001 10:19 AM
> > To: Beowulf (E-mail)
> > Subject: Re: How can I compute the range of signed and unsigned types
> > 
> > Jag wrote : -
> > 
> > > Those sizes are defined for the C language.  In order words, no
> > > matter if you're on a 32-bit machine or a 64-bit machine, an int is
> > > always going to be 32-bit and thus have the same numeric range
> > > because the standards say so.  This goes for all the basic types,
> > > not just int's.
> > 
> > No, the C standard says nothing of the sort.
> > 
> > All the C standard says is that
> > 
> > 1) sizeof (char)  == 1
> > 2) sizeof (short) >= sizeof (char)
> > 3) sizeof (int)   >= sizeof (short)
> > 4) sizeof (long)  >= sizeof (int)
> > 5) sizeof (long long) >= sizeof (long).
> > 
> > It also does not specify that the representation of an int is two's
> > complement, so even on machines with the same sizeof(int) the legal
> > ranges could differ.
> > 
> > -- Jim
> > 
> > James Cownie    <jcownie at etnus.com>
> > Etnus, LLC.     +44 117 9071438
> > http://www.etnus.com
> > 
> > _______________________________________________
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> -- 
> Jared Hodge
> Institute for Advanced Technology
> The University of Texas at Austin
> 3925 W. Braker Lane, Suite 400
> Austin, Texas 78759
> 
> Phone: 512-232-4460
> Fax: 512-471-9096
> Email: Jared_Hodge at iat.utexas.edu
> 
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