Oh F me. Thank you.


Be Appropriate && Follow Your Curiosity

To answer the second part of your question:

Digging into this I tried something "simplier" and got more confused:

perl -e 'print ((.0425*2) + .0025 ) - ((.0425*2) + .0025 ); print $/;'
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just bring the perlrun option -w (warnings) to the rescue:

$ perl -we 'print ((.0425*2) + .0025 ) - ((.0425*2) + .0025 ); print $
+/;'
print (...) interpreted as function at -e line 1.
Useless use of subtraction (-) in void context at -e line 1.
0.0875
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That can be fixed with either extra parens or a plus sign:

$ perl -we 'print (((.0425*2) + .0025 ) - ((.0425*2) + .0025 )); print
+ $/;'
0

$ perl -we 'print +((.0425*2) + .0025 ) - ((.0425*2) + .0025 ); print 
+$/;'
0
[download]

There is a explanation of this in the perldoc for print:

"Be careful not to follow the print keyword with a left parenthesis unless you want the corresponding right parenthesis to terminate the arguments to the print; put parentheses around all arguments (or interpose a + , but that doesn't look as good)."

surprising but true, i get something like, $ perl -we 'my $x = (.0425 * 2) + .0025; print $x - (.042+5 * 2) - .0025, $/;' -9.957 i am using perl 5.8.8

Do you mean that it surprises you that

.0425*2

is different than

.042+5*2

Um... wow... really?

Wise Man Say:   “Floating point numbers like little piles of sand.   Each time you pick one up and move it around, you lose a little sand and pick up a little dirt.”

Except that's not quite true. Each time you pick one up and move it around, you may lose a little sand. "Sand" is only lost when the number has a huge (>53) number of digits in binary.

This usually happens because of the use of numbers that are periodic in binary; they have an infinite number of digits. The OP used 425/10000 and 25/10000 which are both periodic in binary (like 1/3 is in decimal).