floating point precision

2012-01-11 Thread dsmith 
How do you increase floating point precision beyond the default of 6?
example:

double var = exp(-1.987654321123456789);
writeln(var);

-- 0.137016

Assuming this result is only an output format issue and that operations are
still using double's 64 places, if var above is passed to a function, are all
64 places passed?  Must it be passed by reference to make it so?

Re: floating point precision

2012-01-11 Thread Ali Çehreli 

On 01/11/2012 03:51 PM, dsmith wrote:

How do you increase floating point precision beyond the default of 6?
example:

double var = exp(-1.987654321123456789);
writeln(var);

--  0.137016

Assuming this result is only an output format issue and that operations are
still using double's 64 places, if var above is passed to a function, are all
64 places passed?  Must it be passed by reference to make it so?




writefln takes a format argument. The following uses the actual 
precision of the variable but it is wrong as it assumes (knows) that the 
whole part of the result is zero. Since precision is the total of 
significant digits, the digits after the period must be adjusted if the 
whole part is not zero.


import std.math;
import std.stdio;
import std.conv;

double var = exp(-1.987654321123456789);
auto fmt = %. ~ to!string(var.dig) ~ f;
writefln(fmt, var);

Also check out %e, %g, and %a; some of which produce the same 
output depending on the value.


Ali

Re: floating point precision

2012-01-11 Thread Mail Mantis 
All is passed, to print, say, 50 signs after period use following:
writefln(%.50f, var);


2012/1/12 dsmith dsm...@nomail.com:
 How do you increase floating point precision beyond the default of 6?
 example:

 double var = exp(-1.987654321123456789);
 writeln(var);

 -- 0.137016

 Assuming this result is only an output format issue and that operations are
 still using double's 64 places, if var above is passed to a function, are all
 64 places passed?  Must it be passed by reference to make it so?