On 6/12/05, David Abrahams <dave@boost-consulting.com> wrote:
Peder Holt <peder.holt@gmail.com> writes:
<snip>
[Please try to limit the amount of quoted text in your postings]
Ok :)
double< 3,1415927 ,E<+6> >
One more problem: It is not possible to write e.g. the number 3.00011 in the above notation: double_<3,00011> First of all, 00011 is interpreted as 9. Second, there is no way to distinguish this from: double_<3,9> (as I know of) The alternatives would then be: double_<31415927,E<-7> > (which is basically the same as what Andy does in his physical quantities library) DOUBLE(3,00011,E<0>) where 1 is temporarily prepended to 0001 to preserve the preceding zeros. Alternatively: double_<3,DECIMALS(0001)> that has a default value for the exponent.
Not bad. The double type does have 16 decimals precision. Is it an option to use long long for the decimal part?
Only on compilers that support it. Again, that's a non-C++98 extension.
I discovered that using base 2 exponent in stead of base 10 exponent
I don't understand what you mean her. Where is this exponent used? Does the user specify the exponent in base 2 (probably not)? YOu must mean in the representation?
I mean the representation. I just need to find the correct transformation between the E<N> notation and the 2^M notation.
gives you the same accuracy as with a regular double expression.
Is this an absolute requirement, ^^^^
a base 2 exponent or the same accuracy as a double?
this == same accuracy as double
or is an approximation to double sufficient?
I don't know. Better ask your target audience! I'd guess they don't want an approximation, but I'm only guessing.
-- Dave Abrahams Boost Consulting www.boost-consulting.com
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