int 'comparison operator' int # returns the actual result
int 'comparison operator' string # returns an error

Should (5 == '5') (which is "int 'comparison operator' string") return the result ? Or should it throw an error ? Assuming it should return the result, the following may help:

use warnings;

# Write the warnings handler. It only handles runtime errors.
# To have it handle compile-time errors, place it in a 
# BEGIN{} block
$SIG{__WARN__} = sub {
    for my $w (@_) {
       if($w =~ /isn't numeric/) {die "Treated a string as a number"}
       else {print "Warning received: $w"}
    }
};

# Generates a compile-time warning ("...used only once...")
# that the handler doesn't deal with.
$foo = 0;

#Generates a runtime warning ("... uninitialized value ...")
# that the handler *does* deal with.
my $bar;
print $bar, "\n";

$int = 7;
$string = '5foo';

if($int == 7) {print "Yep, \$int == 7\n"}

# Generates a warning that causes the handler to throw
# the error
if($string == 5) {print "Yep, \$string == 5\n"}
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Note, however, that any treatment of $string as a number (eg $string += 3;) will cause the handler sub to throw the error.

Perhaps all you really need is Scalar::Util::looks_like_number() (see 'perldoc Scalar::Util').

use Scalar::Util;
.
.
if(Scalar::Util::looks_like_number($x) &&
   $x > 0) {
   print "It's a positive number - not necessarily an integer";
}
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Maybe even:

use Scalar::Util;
.
.
if(Scalar::Util::looks_like_number($x) &&
   $x > 0 && $x == int($x)) {
   print "It's a positive integer";
}
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But that can be a little dubious, too:

use Scalar::Util;
use warnings;

$string = '5.' . '0' x 30 . '1';;

print $string, "\n";

if(Scalar::Util::looks_like_number($string) &&
  $string > 0 && $string == int($string) ) {
  print "looks like a positive integer\n";
}
else {print "not a positive integer\n"}

__END__
Outputs:
5.0000000000000000000000000000001
looks like a positive integer
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That probably does the right thing imo, but I wouldn't like to guarantee that everyone would agree.

Cheers,
Rob