Bug 366 – Adding trailing zeros to a real literal makes it smaller!
Status
RESOLVED
Resolution
INVALID
Severity
normal
Priority
P2
Component
dmd
Product
D
Version
D1 (retired)
Platform
x86
OS
Windows
Creation time
2006-09-25T10:30:00Z
Last change time
2014-02-15T13:21:07Z
Assigned to
bugzilla
Creator
clugdbug
Comments
Comment #0 by clugdbug — 2006-09-25T10:30:01Z
I know that conversions of decimal literals to floating point is always a bit dodgy -- but this particular behaviour is hard to understand. Happens with lots of numbers.
static assert(
0.6600_0112_2408_4800_3227e-8L <=
0.6600_0112_2408_4800_32270000e-8L);
Comment #1 by bugzilla — 2006-09-25T13:40:11Z
Conversion from decimal to binary is done using "round to nearest", not "truncate towards zero". Round to nearest will 50% of the time result in larger numbers than truncating towards 0. Adding 0's to the literal is the same as truncate towards zero.
Comment #2 by aldacron — 2006-09-26T01:35:16Z
[email protected] wrote:
> http://d.puremagic.com/issues/show_bug.cgi?id=366
>
>
> [email protected] changed:
>
> What |Removed |Added
> ----------------------------------------------------------------------------
> Status|NEW |RESOLVED
> Resolution| |INVALID
>
>
>
>
> ------- Comment #1 from [email protected] 2006-09-25 13:40 -------
> Conversion from decimal to binary is done using "round to nearest", not
> "truncate towards zero". Round to nearest will 50% of the time result in larger
> numbers than truncating towards 0. Adding 0's to the literal is the same as
> truncate towards zero.
In fact, adding ANY digits to the literal makes it smaller. This one
also fails:
static assert(
0.6600_0112_2408_4800_3227e-8L <=
0.6600_0112_2408_4800_3227999999999999999999999999999999e-8L);
Comment #3 by aldacron — 2006-09-26T03:30:16Z
Walter Bright wrote:
> Don Clugston wrote:
>>> Conversion from decimal to binary is done using "round to nearest", not
>>> "truncate towards zero". Round to nearest will 50% of the time result
>>> in larger
>>> numbers than truncating towards 0. Adding 0's to the literal is the
>>> same as
>>> truncate towards zero.
>>
>> In fact, adding ANY digits to the literal makes it smaller. This one
>> also fails:
>>
>> static assert(
>> 0.6600_0112_2408_4800_3227e-8L <=
>> 0.6600_0112_2408_4800_3227999999999999999999999999999999e-8L);
>
> Those extra digits are beyond the precision of the floating point
> format. Try printing them out in %a format to see the bit pattern. The
> behavior is governed by C's strtold, and it does the same thing in gcc.
>
> I can send you the source to strtold if you want to examine it.
Don't worry, I'm a constant user of the %a format!
I've been reading a recent draft of the IEEE 754R standard. It states:
(here μ = real.dig, but could be quadruple.dig if it were expanded).
-----------
If more than μ digits are given for decimal-string to internal
conversions, the result of the conversion shall be
as if it were carried out in two steps: First round the given number to
μ decimal digits, and then convert the
resulting μ-digit number to the target format, in both cases rounding
correctly according to the prevailing
rounding mode.
As a consequence of the foregoing, conversions shall be monotonic:
increasing the value of an internal
floating-point number shall not decrease its value after conversion to
an external character sequence, and increasing the value of a external
character sequence shall not decrease its value after conversion to an
internal floating-point number.
------------
It's the first step (decimal rounding) that appears to be missing.
I don't know how much of a change this is from IEEE 754. Without
monotonicity, you need to be very careful about increasing the precision
of constants.