On Sat, Jun 14, 2003 at 08:56:21AM -0500 deborah wrote:
> How does Perl interpret a string when used in a numeric comparison
> equation?
It depends.
> I accidently used a numeric comparison operator when I was comparing
> strings and I found that no matter what strings I compared, Perl always
> said they were equal.
In that case, both certainly evaluated to the same numerical value. For
instance, this comparison is true:
"14b" == "14c" # in numerical context both are evaluated as 14
this one as well:
"b14" == "c14" # in numberical both are evaluated as 0
And subsequently:
"14b" != "b14" # because 14 != 0
> Any other operation (such as <, >, != ) always proved false, but it was
> always true that 'stringA'=='Bstrg'. Is it just saying that "it is
> true" that stringA and Bstrg are both strings? I thought it would count
> spaces or convert to numbers or something like that. Well, actually, I
> first was surprised that it didn't give me an error since I used the
> wrong type of operator. What is it doing in this case?
It's not an error in Perl. But you get a warning if you 'use warnings'
or employ the -w switch:
perl -we 'print "14b" == "14c"'
Argument "14c" isn't numeric in numeric eq (==) at -e line 1.
Argument "14b" isn't numeric in numeric eq (==) at -e line 1.
1
As a rule of thumb: perl takes the longest numerical prefix of a string
and treats it as number (either integer or float). In the following,
numeric context is enforced by adding 0 to a string:
print "$_"+0, "\n" for "2", "2a", "2.2", "2.2a", "a2", "a2.2";
__END__
2
2
2.2
2.2
But:
print "4e10"+0;
__END__
40000000000
because 4e10 is a number in scientific notation and therefore a valid
number literal.
Tassilo
--
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