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

I learned from some older books, and part of the conventional wisdom was to chomp all input, but I was looking at an example today, and it looks like some kind of "autochomp" is occurring. For instance, in the following code, $data clearly matches 2, but the 2 still has a newline on it, since when I print it without a newline, the newline is there. So why does "$data == 2" match? I would expect the hidden newline to interfere. I would expect "$data =~ /2/" to match; but == matching dosn't seem right to me. What am I missing? 
#!/usr/bin/perl
use strict;
use warnings;

while (my $data = <DATA>) {
#chomp $data; 
print $data;
if ($data == 2){print "->got 2\n"}
}

print "Enter a 2\n";
my $input = <>;
if ($input == 2){print "->got 2\n"}

__DATA__
1
2
3
4

[download]

Replies are listed 'Best First'.
Re: Whatever happened to chomp?
by jsprat (Curate) on Dec 14, 2003 at 14:57 UTC
    2 is a number, and perl is trying to DWYM with the numerical compare (==). Try putting 
    one
two
three
four

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    in __DATA__, and

    change the conditional to if ($data eq '2'). Now see what effect chomp has...

    Update: Struck out the data section and added bit about numerical compare to the first sentence. I shouldn't have included the red herring (one, two, three). The original point was supposed to be that perl will treat a variable like a number in numerical context (== instead of eq).

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      2 is a number.

      No, 2 could be a number or a string base on the context as how you use it. A better answer is to point out what he really misunderstood: the context.

      It is pretty straight to simply change == to eq, and quot 2:

      use Tk;
use Tk::NoteBook;
use strict;
use warnings;
use strict;
use warnings;

while (my $data = <DATA>) {
chomp $data; 
print $data;
if ($data eq "2"){print "->got 2\n"}
}

print "Enter a 2\n";
my $input = <>;
chomp $input;
if ($input eq "2"){print "->got 2\n"}

__DATA__
1
2
3
4

      [download]
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Re: Whatever happened to chomp?
by Limbic~Region (Chancellor) on Dec 14, 2003 at 14:58 UTC
    zentara,
    I am pretty sure this has nothing to do with chomp, but with Perl always trying to DWYM. Reading perldoc perlop I see that == will return true if the two arguments are numerically equal. Consider the following: 
    perl -e 'print "yeah\n" if "p2p" == "f2b"'

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    I am not sure if the behavior is well documented though.

    Cheers - L~R

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      L~R:
      perl -e "use warnings; print qq{yeah$/} if 'p2p' == 'f2b'; print 'p2p'
+ + 0"
Argument "f2b" isn't numeric in numeric eq (==) at -e line 1.
Argument "p2p" isn't numeric in numeric eq (==) at -e line 1.
Argument "p2p" isn't numeric in addition (+) at -e line 1.
yeah
0

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      Non-numeric arguments are treated as 0 in numeric operations. == is a numeric comparision operator, its string equivalent being eq.
      I remember reading somewhere (maybe the camel or llama book) that numbers in string representations may have a newline character after them, which would explain why you don't need to chomp. I couldn't find the passage, though.
      Cheers,
      CombatSquirrel.
      Entropy is the tendency of everything going to hell.
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        You are thinking of '3bar' == '3foo'. It doesn't matter what is after the number, the numeric value of a string with leading digits is the decimal value of those digits. Strings without leading digits are equal to zero.

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        CombatSquirrel,
        The missing warnings and strictures were intentional. I was trying to demonstrate the extent Perl would go to try and DWYM. I guess I should have been more clear - thanks. 
        perl -e 'print "yeah\n" if "3p" == "f2b"'

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        Cheers - L~R
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Re: Whatever happened to chomp?
by derby (Abbot) on Dec 14, 2003 at 15:33 UTC
    As others have pointed out, it is perl DWYM but exactly how does it do that? Well if you look at perl internals, a scalar value can be a string, an integer or a double (or another scalar). By context, it determines which value to use. In a numeric context (==), it will use one of the numeric (integer or double), in a string context, (eq), the string value.

    Check this out: 

    #!/usr/bin/perl
use strict;
use warnings;
use Devel::Peek qw( Dump );

while (my $data = <DATA>) {
chomp $data;
print Dump( $data );
if ($data == 2){print "->got 2\n"}
}

print "Enter a 2\n";
my $input = <>;
print Dump( $input );
if ($input == 2){print "->got 2\n"}

__DATA__
1
2
3
4

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    -derby

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Re: Whatever happened to chomp?
by ysth (Canon) on Dec 14, 2003 at 17:01 UTC
    In numeric context, perl will not warn if a number matches something like: 
    /^(?:
   0\ but\ true  |
   (?:
      \s*    # leading spaces ignored
      [-+]?  # optional sign
      (?:
         (?i: inf|infinity|nan ) | # any case of inf or nan
         (?: \d+ (?:\.\d*)? | \.\d+ ) # digits, optional .
            (?: [eE] [-+]? \d+ )?     # optional exponent
      )
      \s*    # trailing spaces (including "\n")
   )
  )\z
/x;

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    The case you are hitting is the "trailing spaces" being silently ignored.

    If compiled with numeric locale support, either . or the locale's radix will work.

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Re: Whatever happened to chomp?
by Zaxo (Archbishop) on Dec 14, 2003 at 23:02 UTC

    In numeric context, perl accepts numbers in a variable until it runs out of them: 

    $ perl -e' $_ = "123abc"; print 0+$_, $/'
123
$

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    After Compline,
    Zaxo

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