Re: sprintf and decimals

sprintf uses "Banker's rounding". ( Joost points out that this may be compiler-dependant. I'm using ActivePerl. )
4 < 5, so 2.674 rounds to 2.67
6 > 5, so 2.676 rounds to 2.68
5 = 5, and 7 is odd, so 2.675 rounds to 2.67
5 = 5, and 8 is even, so 2.685 rounds to 2.69

Banker's rounding (the even/odd bit) helps in the case of where one is adding a lot of rounded .5's

>perl -e "print 10.5 + 15.5 + 25.5"
51.5 (Unrounded)
>perl -e "print 11 + 16 + 26"
53 (Conventional, delta = 1.5)
>perl -e "print 10 + 16 + 26"
52 (Banker's, delta = 0.5)
[download]

Of course, it doesn't help in the case where one is adding lots of rounded .6's, but it doesn't hurt either.

>perl -e "print 10.6 + 15.6 + 25.6"
51.8 (Unrounded)
>perl -e "print 11 + 16 + 26"
53 (Conventional, delta = 1.2)
>perl -e "print 11 + 16 + 26"
53 (Banker's, delta = 1.2)
[download]

While it can produce a different result when the data is "well" distributed, it doesn't hurt there either.

>perl -e "print 0.0+0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9"
4.5 (Unrounded)
>perl -e "print 0+0+0+0+0+1+1+1+1+1"
5 (Conventional, delta = 0.5)
>perl -e "print 0+0+0+0+0+0+1+1+1+1"
4 (Banker's, delta = 0.5)
[download]

Update: Added background on banker's rounding.

Comment on Re: sprintf and decimals Select or Download Code

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Re^2: sprintf and decimals by Joost (Canon) on Nov 15, 2006 at 18:54 UTC
sprintf uses "Banker's rounding" Does it always? The docs appear to say it uses the system's sprintf() for floating points, and as far as I know that isn't guaranteed to do banker's rounding. (See for example the remarks here) "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply]
Re^3: sprintf and decimals by ikegami (Patriarch) on Nov 15, 2006 at 19:27 UTC
What about `sprintf` in POSIX? Is that one compiler-dependant too?	[reply] [d/l]
Re^4: sprintf and decimals by Joost (Canon) on Nov 15, 2006 at 19:32 UTC
Interestingly, this is the implementation of POSIX::sprintf in perl 5.8.7: `sub sprintf { usage "sprintf(pattern,args)" if @_ == 0; CORE::sprintf(shift,@_); }` [download] So I would say "yes" :-) "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l]
Re^2: sprintf and decimals (theory) by tye (Sage) on Nov 15, 2006 at 23:16 UTC
5 = 5, and 7 is odd, so 2.675 rounds to 2.67 5 = 5, and 8 is even, so 2.685 rounds to 2.69 Well, in theory. But floating point numbers can't actually represent 2.675 nor 2.685, so the rounding for these cases is actually determined by what numbers can be represented in floating point and are closest to those numbers: `DB> x sprintf "%.20f %.2f", (2.675)x2 0 '2.67499999999999980000 2.67' DB> x sprintf "%.20f %.2f", (2.685)x2 0 '2.68500000000000010000 2.69' DB>` [download] which happens to agree with your assertion, but for the wrong reason. Change 6 to 4 to see a case that disagrees (at least on my system): `DB> x sprintf "%.20f %.2f", (2.475)x2 0 '2.47500000000000010000 2.48' DB> x sprintf "%.20f %.2f", (2.485)x2 0 '2.48499999999999990000 2.48' DB>` [download] Finally, just to show how close floating point can get on the other side of one of these examples: `DB> x pack "d", 2.675 0 "ffffff\cE\@" DB> x unpack "d", "gfffff\cE\@" 0 2.675 DB> x sprintf "%.20f %.2f", (2.675)x2 0 '2.67499999999999980000 2.67' DB> x sprintf "%.20f %.2f", (unpack "d", "gfffff\cE\@")x2 0 '2.67500000000000030000 2.68'` [download] And, this may give you some idea how many different values you might compute that Perl would report as just "2.675" but that can round differently: `DB> x unpack "d", "qfffff\cE\@" 0 2.675 DB> x unpack "d", "rfffff\cE\@" 0 2.67500000000001` [download] - tye	[reply] [d/l] [select]
Re^2: sprintf and decimals by nosbod (Scribe) on Nov 15, 2006 at 18:46 UTC
thanks. So that'll be why this bit of code returns the following `#!/usr/bin/perl $w=2.315; $x=2.3150141; $y=sprintf("%.2f", $w); $z=sprintf("%.2f", $x); print "$w rounded is $y\n"; print "$x rounded is $z\n";` [download] 2.315 rounded is 2.31 2.3150141 rounded is 2.32	[reply] [d/l]
Re^3: sprintf and decimals by ikegami (Patriarch) on Nov 15, 2006 at 18:55 UTC
oh! I didn't know it looks at more than the next digit. To continue the example in my previous post, that means 5.1 > 5, so 2.6751 rounds to 2.68	[reply]