Re: Scheme help...

[Earl Hood <ehood@xxxxxxxxxxxxxxxxx>]

> Anyhow, the question is about the difference between this:
> 
> (define foo (bar))
> and
> (define (foo) (bar))
> 
> I don't have my Perl book with me, but they are roughly analogous to:
> 
> foo = bar()

$foo = bar();

> foo =  function (){ # no args
>     return bar();
> }

sub foo { bar(); }

Of course, it is not exact with Scheme due the inherent difference
in the languages.  But I believe closely parallels Paul's description
about (define ...).  In the Perl statements, "foo" is a different
in each statement.  The first, it is a scalar variable, and the second, it
is a routine.

In the first, bar is invoked and the return value is assigned to $foo.
I.e. Using $foo will not call bar again.  In the second, every call to
foo will call bar.  Hence the return value of foo may not always be the
same depending on the behavior of bar.

	--ewh


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