dynamic scoped variables in the debugger

[Nathan]

Hi,

I'm running Perl v5.8.8 compiled for linux. I was reading about the perl
debugger and ran the following script with the command "perl -d":

1 #!/ usr/bin/perl
2 $_ = "foo";
3 /foo/;
4
5 print "$&";
6 print "\n";

As expected it prints "foo" before the newline, and if I check the value
of $& in the debugger it does show "foo". However, if I omit line 5 of
the program then the debugger says $& is undefined when it reaches the
last print statement. I don't understand how this can be, is there some
subtlety to the way the debugger treats dynamically scoped variables?
Any insight anyone can offer would be great, thanks.

-Nathan

 

[xhoster]

Hi,

I'm running Perl v5.8.8 compiled for linux. I was reading about the perl
debugger and ran the following script with the command "perl -d":

1 #!/ usr/bin/perl
2 $_ = "foo";
3 /foo/;
4
5 print "$&";
6 print "\n";

As expected it prints "foo" before the newline, and if I check the value
of $& in the debugger it does show "foo". However, if I omit line 5 of
the program then the debugger says $& is undefined when it reaches the
last print statement.

When the program is compiled, if the compiler sees $& being used,
then it tuns on the (expensive) code which causes $& to be set during
regex operations. If $& is not seen, then this code is not activated
and $& does not get populated.

I don't understand how this can be, is there some
subtlety to the way the debugger treats dynamically scoped variables?

This is not related to the debugger itself. The debugger is just giving
you a way arrange things so that you can inspect $& without perl knowing
you are going to do so. You can get the same behaviour using string eval:

perl -le '$_="foo"; /foo/; eval q{print "$&"}'

Also, $& is not just a dynamic variable (like $foo::bar), but a special
variable. It is the implementation details of this specialness that is
tripping you up.

Xho

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[Nathan]

When the program is compiled, if the compiler sees $& being used,
then it tuns on the (expensive) code which causes $& to be set during
regex operations. If $& is not seen, then this code is not activated
and $& does not get populated.


This is not related to the debugger itself. The debugger is just giving
you a way arrange things so that you can inspect $& without perl knowing
you are going to do so. You can get the same behaviour using string eval:

perl -le '$_="foo"; /foo/; eval q{print "$&"}'

Also, $& is not just a dynamic variable (like $foo::bar), but a special
variable. It is the implementation details of this specialness that is
tripping you up.

Xho

Thanks Xho, I vaguely suspected something along those lines, but I guess
I underestimated how "special" the special variables are. Perl gives you
everything you need except consistency.

-Nathan