> $function_used isnt being treated as a variable (as you know)

Nope, I doubt this! Even without being able to test the code you posted. (did you?)

memcpy should be deleted now.

My best guess is that in your real code $function_used doesn't contain what you think it should.

Maybe a problem with scoping?

> So how can one solve this? !!!!!!!!!!!! I solved this by removing the global identifier. !!!!!!!!!!!!

Your keyboard is broken, maybe that's the problem?(...??????????)

:)

Cheers Rolf

(addicted to the Perl Programming Language)

update

now I can test, the code you posted explicitly informs you about the wrong semicolon: 

> perl
$function_used = "memcpy";
$function_call = "memcpy(a, b, sizeof(a));"
$function_call =~ s/$function_used//g;
Scalar found where operator expected at - line 3, near "$function_call
+"
    (Missing semicolon on previous line?)
syntax error at - line 3, near "$function_call "
Execution of - aborted due to compilation errors.

[download]

and the corrected code doesn't have any of your imagined "problems"

> perl
$function_used = "memcpy";
$function_call = "memcpy(a, b, sizeof(a))";
$function_call =~ s/$function_used//g;
print $function_call;
__END__
(a, b, sizeof(a))

[download]

so I think you should better take a look into How (Not) To Ask A Question :)

In reply to Re: Treating a variable as a variable in a regular expression? by LanX
in thread Treating a variable as a variable in a regular expression? by parkinglot

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