OpenEuphoria

Re: Forward Definitions

• Posted by Lucius Hilley III <lhilley at CDC.NET> Feb 25, 1999
• 389 views

Thank you David.  This is a WONDERFUL example of how routine_id() does
and doesn't work.  If only you had shown how actually using the ID
works we would have PERFECT example documentation on how you can and
can't use routine_id.

    Lucius L. Hilley III

On Wed, 24 Feb 1999 13:00:19 -0800, Cuny, David <David.Cuny at DSS.CA.GOV>
wrote:

>David Guy wrote:
>
>>>3. Change the scope of routine_id to the run time scope, not compile
time.
>> ...
>> If it's not too much trouble, could you give a simple example of what you
>> mean by the above statement, and how/why it would/could be a problem?
>
>routine_id returns an identifier to a user-defined routine. It will fail
>(return -1) under one of the following circumstances:
>
>1. The routine is not defined, or local to another include file;
>2. The routine is not part of the current scope; or
>3. The routine is a built-in, such as gets or puts.
>
>routine_id can only "see" routines that have been defined up to the routine
>that it is embedded into. For example:
>
>   procedure foo()
>   end procedure
>
>   procedure test( sequence procname )
>        if routine_id( procname ) = -1 then
>         puts( 1, "can't see " & procname & "\n" )
>      else
>         puts( 1, "can see " & procname & "\n" )
>      end if
>   end procedure
>
>   procedure bar()
>   end procedure
>
>   test( "puts" )
>   test( "foo" )
>   test( "bar" )
>
>if you run this code, you will get the following results:
>
>   can't see puts
>   can see foo
>   can't see bar
>
>'puts' cannot be seen because it is a built-in function. This behavior
>certainly suprised me when I first saw it, but I suspect that there are
>compelling reasons for being that way. My guess is that 'routine_id' is
>probably pretty gruesomely coded, and we are lucky to have *anything* like
>it at all.
>
>'foo' can be seen, because it was defined before 'test'. Since 'test' can
>see 'foo', so can 'routine_id'.
>
>'bar' cannot be seen, because it was defined after 'test'. Since 'test'
>cannot see 'foo', neither can 'routine_id'.
>
>The way I would *prefer* the code to work is:
>
>   'bar' can be seen, because it is *called* after 'bar' is defined.
>
>There are various reasons for the preference; most of them can be worked
>around by using routine_id at the point were I need the routine id.
>
>-- David Cuny