Passing file-handle to a perl routine

Question

Please Unix guys/gals:
I would like to pass a file-handle to a perl routine. In faq5 I've found that wherever perl expects a file-handle, one can supply an integer containing it.
As far as I can see both
open (handle, ".filename"); $hand = *handle ;
and
open($hand,">filename) ;
produce an integer $hand that contains the handle.
Now I can pass this to a routine:
routine $hand ;
and I can use it in
while (<$hand>) { ... }
or
print $hand "whatever\n" ;
and it is ok. The problem is with this structure:
print $hand << "label" ;
this an' that.....
label
Well, in the main script I can use
print handle << "label ";
whilst using $hand in other constructs, but handle is not known in the
routine. Could you tell me where I got it wrong or how could I work around
this feature, if it is a feature.
Thank you,
Miklos

Answer 1

The problem is with this structure:
print $hand << "label" ;
The problem with that non-working example is one of syntax and operator precedence. The << is seen as the binary-shift-left operator: you're operating on the $hand, which uses up the "label" parameter. Now you've got a corrupted filehandle. You also have only a single parameter in the print function, so that value is perhaps actually printed instead of being used as a filehandle. Because the <<"label" was not seen as a multiline string thing, the print command is now done, and the string is treated as actual perl.
What to do? The print function is prototyped to take an array of stuff to print, so place the stuff you want printed in parens:<pre> print $hand (<< "label");
blah
blah
label</pre>
To see the difference in how perl interprets the two syntaxes, run them with perl -MO=Deparse