That's not a bug in Perl. It's an artifact of using floating point numbers. There are many numbers that cannot be represented exactly in floating point.

At a glance, it appears that none of those would be exactly represented in the IEEE floating point formats. (I could be wrong.) So there would be some error in the calculation in any language, on any processor.

Strictly, it's not because it's floating point, it's because it's binary floating point.

When decimal 1.51 is converted to a binary fraction the result is:

  1.1000_0010_1000_1111_0101_1100_0010_1000_1111_0101_1100_0010_1000_1111_0101_1100_0010_1000_1111_0101...

which you can see is a repeating binary fraction.

As you say, most decimal fractions are like this... so before any errors can be introduced by rounding and what not, most decimal fractions have a built-in "representation" error.

Most of the time you won't see the "representation" error, because conversion back to decimal rounds it off. This can lead to bafflement when two values look the same when printed out, but fail $x == $y. Addition and subtraction are the more difficult floating point operations, so you're more likely to see the problems there. Consider:

  print 1.09 - 1, "\n" ;
  print "0.84 - 0.34 == ", 0.84 - 0.34, 
       ( 0.84 - 0.34 == 0.5 ? " ==" : " but !=" ), " 0.5\n" ;
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which gives:

  0.0900000000000001
  0.84 - 0.34 == 0.5 but != 0.5

it really makes me wonder why we persist in using binary floating point for decimal arithmetic !

This issue is caused by binary floating point. But, it is an artifact of any floating point notation. For example,

1/3 + 1/3 + 1/3 != 1
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for any finite number of digits using decimal representations.

Given any base, there will be some fractions that cannot be properly represented and would therefore cause this problem. (At least until such time as we have infinite precision floating point representations.<grin/>)

G. Wade