How can I compute the range of signed and unsigned types

Walter B. Ligon III walt at clemson.edu
Wed Apr 18 12:57:05 EDT 2001


--------

Um, I'm sorry but the information given below is wrong.

for an unsigned integer of b bits the range is

	0 to 2^b - 1

for a signed integer of b bits using 2's complement, the range is

	-(2^(b-1)) to 2^(b-1)-1

for example, for 4 bit 2's comp binary numbers, the range is

	-2^3 to 2^3-1 or -8 to 7

for a signed integer of b bits using 1's complement or sign magnitude, the 
range is

	-(2^(b-1)-1) to 2^(b-1)-1

and there are two representations of 0 (0 and -0).  Thus for the above
example the range would be

	-2^3-1 to 2^3-1 or -7 to 7

As pointed out, FP is more complex - you have to under stand the range
of the mantissa (which is basicly in sign magnitude) and the range of the
exponent (which is in something called "excess notation" which is yet
another integer format with yet again different range properties).

Again, I would suggest the read consult a standard text on the subject.
Hennessy and Patterson's architecture book is a reasonable one.

Walt

> 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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-- 
Dr. Walter B. Ligon III
Associate Professor
ECE Department
Clemson University



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