How to fix 'Bareword found' issue in perl eval()-CodePudding

The following code returns "Bareword found where operator expected at (eval 1) line 1, near "*,out" (Missing operator before out?)"

$val = 0;
$name = "abc";
$myStr = '$val = ($name =~ in.*,out [)';
eval($myStr);

As per my understanding, I can resolve this issue by wrapping "in.*,out [" block with '//'s.

But that "in.*,out [" can be varied. (eg: user inputs). and users may miss giving '//'s. therefore, is there any other way to handle this issue.? (eg : return 0 if eval() is trying to return that 'Bareword found where ...')

CodePudding user response：

The magic of (string) eval -- and the danger -- is that it turns a heap of dummy characters into code, compiles and runs it. So can one then use '$x = ,hi'? Well, no, of course, when that string is considered code then that's a loose comma operator there, a syntax eror; and a "bareword" hi.^† The string must yield valid code

In a string eval, the value of the expression (which is itself determined within scalar context) is first parsed, and if there were no errors, executed as a block within the lexical context of the current Perl program.

So that string in the question as it stands would be just (badly) invalid code, which won't compile, period. If the in.*,out [ part of the string is in quotes of some sort, then that is legitimate and the =~ operator will take it as a pattern and you have a regex. But then of course why not use regex's normal pattern delimiters, like // (or m{}, etc).

And whichever way that string gets acquired it'll be in a variable, no? So you can have /$input/ in the eval and populate that $input beforehand.

But, above all, are you certain that there is no other way? There always is. The string-eval is complex and tricky and hard to use right and nigh impossible to justify -- and dangerous. It runs arbitrary code! That can break things badly even without any bad intent.

I'd strongly suggest to consider other solutions.

^† A problem if you're into warnings, and we all are.

CodePudding user response：

The following isn't valid Perl code:

$val = ($name =~ in.*,out [)

You want the following:

$val = $name =~ /in.*,out \[/

(The parens weren't harmful, but didn't help either.)

If the pattern is user-supplied, you can use the following:

$val = $name =~ /$pattern/

(No eval EXPR needed!)

Note from the correction that the pattern in the question isn't correct. You can catch such errors using eval BLOCK

eval { $val = $name =~ /$pattern/ };
die("Bad pattern \"$pattern\" provided: $@") if $@;

A note about user-provided patterns: The above won't let the user execute arbitrary code, but it won't protect you from patterns that would take longer than the lifespan of the universe to complete.