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

So I want to create an if statement that will not allow the input of a non-integer into STDIN. I tried two things that did not work:

The first: 


if ($numind!=int) {
    print "**This is an invalid entry!", "\n", "Please try again.**\n"
+;
    goto START;
    }

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The second: 

if ($numind!=int(1..10000)) {
    print "**This is an invalid entry!", "\n", "Please try again.**\n"
+;
    goto START;
    }

[download]

The first error states that I have an uninitialized value, which makes sense but the second one, instead of limiting a decimal, limits all input, integer or not, and prints the "Invalid entry" output.

Any help is appreciated,

Melderon

Replies are listed 'Best First'.
Re: using int and the range function
by toolic (Bishop) on Jan 15, 2015 at 21:40 UTC
    One way using \D: 
    use warnings;
use strict;

while (<DATA>) {
    chomp;
    print "This is an invalid entry: $_\n" if /\D/;
}

__DATA__
555
3
1234
11-55
fadf
000
-402
1 2
1.2

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Re: using int and the range function
by AnomalousMonk (Archbishop) on Jan 15, 2015 at 21:48 UTC
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Re: using int and the range function
by Laurent_R (Canon) on Jan 15, 2015 at 22:42 UTC
    You could also use something like this: 
    print "This is an integer\n" if int $input eq $input;

    [download]
    Please note that it might at first glance seem more appropriate to use the numeric comparison operator ==, but this is not the case here. With the eq operator, you will detect that "4y" is not an integer, while with the == operator will silently do the conversion into an integer and report a wrong result.

    Having said that, I would probably tend to prefer some of the regexes already mentioned such as /^\d+$/ or, used negatively, /\D/.

    Je suis Charlie.
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      I think this is closest to the OP's intent, because it also detects if the number doesn't fit into an "integer".

      But I'd change the if condition to 
      ... if $input eq int $input;

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      because for humans that's easier to parse than 
      ... if int $input eq $input;

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        Yes, you are right, it makes precedence comprehension a little bit easier. I actually thought of changing it the way you put it when I posted it, but since I tested it under the debugger the way I wrote it, I prefered not to change my post to something untested. In a real program, I would probably change it to make it clearer.
        Je suis Charlie.
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Re: using int and the range function
by pme (Monsignor) on Jan 15, 2015 at 21:32 UTC
    You can check that using regexp: 
    if ($numind !~ /^\d+/) {
...
}

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    Update: regexp is anchored to the beginning of the string.
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[d/l]
      c:\@Work\Perl>perl -wMstrict -le
"my $numind = '345xy';
 if ($numind !~ /^\d+/) { die qq{'$numind' is not a number}; }
 print 'sum is ', 123 + $numind;
"
Argument "345xy" isn't numeric in addition (+) at -e line 1.
sum is 468

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      Give a man a fish:  <%-(-(-(-<

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      Also need to make sure there is nothing after the digits:

      while (<>)
{
    chomp;
    if (/^\d+$/)
    {
         print "$_ is an integer\n";
    }
}

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[d/l]
        We would also take into account sign or '0x' prefix for hexadecimal numbers. This way we will end up reimplementing Scalar::Util::looks_like_number() what AnomalousMonk suggested below.
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Re: using int and the range function
by Anonymous Monk on Jan 15, 2015 at 22:19 UTC
    Should't it also print "argument isn't numeric in int" for the second one?
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