Re: Please help me understand string eval better

Your main problem is that perl does not consider '09' to be a number. The function 'eval' will fail no matter what you do. Why not explicitly remove the leading zero?

use strict;
use warnings;
use Scalar::Util qw(looks_like_number isdual);
my $transf = '/3+6';
my $constant = '09';
my $var = \$constant;
#if (looks_like_number($$var)) {
if (!looks_like_number($$var)) {         #correction per [haukex]
    print "$$var is not a number\n";
}
else {
    print "$$var looks like a number\n";
}

if (!isdual($$var)) {
    print "$$var does not have a hidden numeric value\n";
} 


$$var =~ s/^0*//;   # make it into a valid number

my$expr = "$$var$transf";
my $ans = eval "$$var$transf";
print "$expr yields: ";
if (!$@) { 
    print $ans, "\n";
}
else {
    print $@, "\n";
}
[download]

Bill

Comment on Re: Please help me understand string eval better Download Code

Replies are listed 'Best First'.
Re^2: Please help me understand string eval better by haukex (Archbishop) on Jun 13, 2017 at 15:00 UTC
perl does not consider '09' to be a number Perl does consider the string `"09"` to be a number, `0 + "09" == 9`. The only question is, does the OP want this to be interpreted as an octal, which in this case would be invalid, or not. `$ perl -wMstrict -le 'print 77' 77 $ perl -wMstrict -le 'print 077 ' 63 $ perl -wMstrict -le 'print 0 + "077"' 77 $ perl -wMstrict -le 'print oct "077"' 63 $ perl -wMstrict -le 'print eval " 77"' 77 $ perl -wMstrict -le 'print eval "077"' 63` [download] `if (looks_like_number($$var)) { print "$$var is not a number\n";` [download] You've got mistake in your logic there. Also note that `looks_like_number` tells you whether the string will properly auto-convert to a number (`0+$str`), but does not cover all of the numeric literals that Perl accepts in its source code, i.e. what eval will parse, such as `0b1001` or `1234_56`; strings like `"0b1001"` won't automagically convert to numbers. I wrote about this a bit more at Re^3: why are hex values not numbers?	[reply] [d/l] [select]

Replies are listed 'Best First'.

Re^2: Please help me understand string eval better
by haukex (Archbishop) on Jun 13, 2017 at 15:00 UTC

perl does not consider '09' to be a number

Perl does consider the string "09" to be a number, 0 + "09" == 9. The only question is, does the OP want this to be interpreted as an octal, which in this case would be invalid, or not.

$ perl -wMstrict -le 'print        77'
77
$ perl -wMstrict -le 'print       077 '
63
$ perl -wMstrict -le 'print  0 + "077"'
77
$ perl -wMstrict -le 'print  oct "077"'
63
$ perl -wMstrict -le 'print eval " 77"'
77
$ perl -wMstrict -le 'print eval "077"'
63
[download]

if (looks_like_number($$var)) { print "$$var is not a number\n";
[download]

You've got mistake in your logic there. Also note that looks_like_number tells you whether the string will properly auto-convert to a number (0+$str), but does not cover all of the numeric literals that Perl accepts in its source code, i.e. what eval will parse, such as 0b1001 or 1234_56; strings like "0b1001" won't automagically convert to numbers. I wrote about this a bit more at Re^3: why are hex values not numbers?

[reply]
[d/l]
[select]