log() and int() problem

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

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Re: log() and int() problem by moritz (Cardinal) on Dec 25, 2012 at 17:02 UTC
log returns a floating-point number, so the results are prone to floating-point inaccuracies: `$ perl -wE 'printf "%.20f\n", log(125)/log(5)' 3.00000000000000044409` [download] A safer test is something like this here, which only uses integers: `use strict; use warnings; use 5.010; sub is_power_of { my ($num, $base) = @_; while ($num > 1) { return 0 if $num % $base != 0; $num /= $base; } return 1; } say is_power_of(125, 5);` [download] Perl 6 - the future is here, just unevenly distributed	[reply] [d/l] [select]
Re^2: log() and int() problem by Anonymous Monk on Dec 25, 2012 at 21:35 UTC
Idly curious why a loop and "/=" are needed. Not questioning but curious. Example?	[reply]
Re^3: log() and int() problem by Anonymous Monk on Dec 25, 2012 at 23:17 UTC
Add `print "$num\n";` and you can see how the loop works	[reply] [d/l]
Re: log() and int() problem by toolic (Bishop) on Dec 25, 2012 at 17:02 UTC
log returns a floating point value, not an integer. See this FAQ	[reply]
Re: log() and int() problem by LanX (Saint) on Dec 25, 2012 at 17:17 UTC
when using Data::Dumper you will see that $l is seen as a string not an integer `print "LOG == ",$l,Dumper \$l,"\n";` prints `LOG == 3$VAR1 = \'3';` As others pointed out log only returns floats up to certain precision, which IIRC also depends on the current implementation of Perl (that is compiling options). ~~And internally floats are stored in the string-slot of a variable.~~ That means the slight calculation error might be too small to be shown by a print but is still sufficient to let `==` fail. Even if print would show the discrepancy, your approach wouldn't always work. So better go the other way round and check if `125 == 5**$x` with `$x=int($l+$tolerance)` EDIT: The approach Moritz updated into his post is even better. Cheers Rolf	[reply] [d/l] [select]
Re^2: log() and int() problem by dave_the_m (Monsignor) on Dec 25, 2012 at 18:34 UTC
And internally floats are stored in the string-slot of a variable No they're not. An xpvnv has separate slots for an int, a float and a string: `[davem@pigeon bleed]$ p -MDevel::Peek -e'$x=1; $x=2.2; $x = "three"; D +ump $x' SV = PVNV(0x1ae3150) at 0x1b021d0 REFCNT = 1 FLAGS = (POK,pPOK) IV = 1 NV = 2.2 PV = 0x1afb360 "three"\0 CUR = 5 LEN = 16` [download] Dave.	[reply] [d/l]
Re^3: log() and int() problem by LanX (Saint) on Dec 25, 2012 at 18:49 UTC
OK, thanks for correcting! (I already suspected that I should place one more "IIRC". :) Maybe I remembered something different or Scalar::Util::dualvar confused me or the reality in perl is even more complex. (and the latter wouldn't surprise me) Cheers Rolf PS: or was it JS ... ?!? UPDATE: For the records, here the source of my misunderstanding from perlnumber: Perl can internally represent numbers in 3 different ways: as native integers, as native floating point numbers, and as decimal strings. Decimal strings may have an exponential notation part, as in "12.34e-56" . Native here means "a format supported by the C compiler which was used to build perl". emphasis added.	[reply]
Re^2: log() and int() problem by Anonymous Monk on Dec 25, 2012 at 21:03 UTC
Thank you very much. I solved my issue with adding and subtracting 1 from the returned value. I think that perl should do this internally. `my $l = log(125) / log(5); $l += 1; $l -= 1; print $l == int($l);` [download] Thank you all for the great answers.	[reply] [d/l]
Re^3: log() and int() problem by eyepopslikeamosquito (Archbishop) on Dec 25, 2012 at 22:40 UTC
I solved my issue with adding and subtracting 1 from the returned value Sorry, but that's a terrible "solution". Your example floating point inaccuracy happens to be slightly greater than three. Since the int function truncates, consider what happens if the inaccuracy happens to be slightly less than three. Though a crude fix would be to add 0.5 (i.e. `int($l+0.5)` instead of `int($l)`), the perl documentation advises against using int for rounding and suggests sprintf and the POSIX `floor` and `ceil` functions as sounder alternatives. Still don't understand why you don't go with moritz's solution.	[reply] [d/l] [select]
Re^4: log() and int() problem by LanX (Saint) on Dec 25, 2012 at 23:19 UTC
Re^3: log() and int() problem by LanX (Saint) on Dec 25, 2012 at 21:11 UTC
> I solved my issue you're very optimistic. I delved into Devel::Peek and I'm still not sure which type casting or crossed thresholds produced different results for 2 and 3. Anyway you're voyaging dangerous grounds, don't be surprised if your Float to Int conversions breaks again. Better rely on on already shown pure integer arithmetic for your integer (sic) tests. (rule of thumb!) Cheers Rolf PS: lanx@nc10-ubuntu:~$ perl -MDevel::Peek -e'$\|=1;$e=2;$x=log(5$e)/log( +5); Dump $x; print ("\n", ($x==$e) ? "" : "not " ,"equal\n\n"); Dump + $x;' SV = NV(0x9c86840) at 0x9c70f68 REFCNT = 1 FLAGS = (NOK,pNOK) NV = 2 equal SV = PVNV(0x9c50a60) at 0x9c70f68 REFCNT = 1 FLAGS = (IOK,NOK,pIOK,pNOK) IV = 2 NV = 2 PV = 0 lanx@nc10-ubuntu:~$ perl -MDevel::Peek -e'$\|=1;$e=3;$x=log(5$e)/log( +5); Dump $x; print ("\n", ($x==$e) ? "" : "not " ,"equal\n\n"); Dump + $x;' SV = NV(0x8fde840) at 0x8fc8f68 REFCNT = 1 FLAGS = (NOK,pNOK) NV = 3 not equal SV = PVNV(0x8fa8a60) at 0x8fc8f68 REFCNT = 1 FLAGS = (NOK,pIOK,pNOK) IV = 3 NV = 3 PV = 0 [download] UPDATE: improved code	[reply] [d/l]