I thought that the 'mp' in mpfr stood for multi-precision?

Aaah ... but when a double is assigned to an X-bit (where X > 53) precision mpfr object, the first 53 bits are set the same as the double, and the following bits are all 0.
(To get an X-bit representation of (eg) 1.3, you would need to assign the string '1.3', not the double 1.3.)
A small demo: 
use strict;
use warnings;
use Math::MPFR qw(:mpfr);

my $prec_53   = Rmpfr_init2(53);
my $prec_4000 = Rmpfr_init2(4000);

my $d = 1.3;

Rmpfr_set_d($prec_53, $d, MPFR_RNDN);
# The 53 bits of $prec_53 are now set to:
# 10100110011001100110011001100110011001100110011001101

Rmpfr_set_d($prec_4000, $d, MPFR_RNDN);
# The first 53 bits are as for $prec_53
# The remaining 3947 bits are 0.

# But we'll run a check:

die "Things aint right"
 if $prec_53 != $prec_4000;

print "$prec_53\n";
# prints 1.3

print "$prec_4000\n";
# prints 1.3000000000000000444089209850062616169452667236328125

[download]
So if I want to assign (from a string) the value held by that double to a 53-bit precision object I can just assign the string "1.3".
But if I want to assign (from a string) the same value to a 4000-bit object then I need to assign the latter string.

I assume that the latter value is an exact representation of the value of the double because after the 53rd decimal digit, all (1153) following decimal digits are 0.

Cheers,
Rob

In reply to Re^9: subtraction issue by syphilis
in thread subtraction issue by Anonymous Monk

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