RE: [Dev-C++] bit extraction problem

[Ioannis V. <iv...@em...>]

One "analog" approach that I can think is to convert the
numeric_limits<long>::max() and min() to strings and count the digits. 

Then compare the number of digits of the input with the number of digits of
max or min depending if the input has a minus sign.

If the number if digits is equal  with the equivalent limit, compare if the
first digit is larger/smaller of the corresponding limit (depending on the
minus sign). And so on.






Ioannis Vranos
 
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> -----Original Message-----
> From: Per Westermark [mailto:pw...@ia...] 
> Sent: Sunday, August 17, 2003 4:19 PM
> To: Ioannis Vranos
> Cc: 'Dev C++'
> Subject: RE: [Dev-C++] bit extraction problem
> 
> 
> One alternative is to count the number of digits, and take the log.
> 
> One decimal digit consumes 1/log(2) binary digits,
> or approx 3.322 bits. However, it is still only possible to 
> get close to
> the true limit - since the log of a number needs at least as many
> value-digits as the number for an exact result :-)
> 
> Another approach - which is exact - is to perform the decimal->binary
> conversion in own code, and for each added decimal digit check if the
> binary number overflows. For a positive number decimal input 
> number, the
> resulting binary should increase for each iteration. If the 
> input number
> is negative, it is simple to check for overflow by negating the result
> twice and check that the original value was returned.
> 
> For a traditional machine with two-complement handling, a 
> negation at the
> end will not overflow. However, I'm not sure if that is true for all
> strange machines out there, that uses other ways of encoding negative
> numbers. Maybe it is possible to negate a number twice and check if
> original value is returned. Anyway, right now, 2-complement rulez :-)