Re: When is a 2 not a 2?

This is real (pardon the pun) interesting. I thought: they have missed the obvious, use int. Wrong:

my $x;

for ($x=1; int($x) <= 2; $x += 0.1) {
  print "$x\n";
}

print "Stop: x=$x\n";
print "No, x is not 2.\n" if $x != 2;
[download]

Gives:

1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
Stop: x=3
No, x is not 2.
[download]

So then I RTFM on int.
Changing the comparisons to text (the comparison operators stringify both sides) the result is correct, but of course this is not a solution for numbers greater than 9.

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Re^2: When is a 2 not a 2? by ikegami (Patriarch) on Jan 31, 2008 at 10:24 UTC
int(2.1) is 2, which is <= 2, so the loop won't end then like it should. This is a completely different bug than the OP's. You were maybe thinking something more along the lines of `int($x10)/10 <= 2`, but that would only work with some combination of numbers. `int($x10+0.5)/10 <= 2` might be ok, but you'd still be needlessly accumulating error in your variable. As already mentioned (although not in so many words), if you want to loop a certain number of times, count the number of loops passes and derive the decimal number from the loop pass number.	[reply] [d/l] [select]

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Re^2: When is a 2 not a 2?
by ikegami (Patriarch) on Jan 31, 2008 at 10:24 UTC

int(2.1) is 2, which is <= 2, so the loop won't end then like it should. This is a completely different bug than the OP's.

You were maybe thinking something more along the lines of int($x*10)/10 <= 2, but that would only work with some combination of numbers. int($x*10+0.5)/10 <= 2 might be ok, but you'd still be needlessly accumulating error in your variable.

As already mentioned (although not in so many words), if you want to loop a certain number of times, count the number of loops passes and derive the decimal number from the loop pass number.

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