On Mon, Mar 15, 2010 at 5:36 PM, Michael Bevilacqua-Linn
<michael.bevilacqualinn@gmail.com> wrote:
A first-class function (or procedure) means that there's a facility in the language that allows you to create procedures at execution time, store a reference to them in a variable, basically treat them as you would any other piece of data that your program is working with.
No. This is wrong. As you can have closures in a compiled language. In a compiled language, closures might be created by doing lambda lifting, which creates a named function with extra parameters for the variables that used to be free.
In other words, a function such as:
(define (make-counter x)
(lambda (n)
(begin
(setf! x (+ x n))
x)))
Might compile to the following procedures (illustrative only):
(define (lifted-anonymous-function-1 n FREE_X)
(begin
(set-box! FREE_X (+ (unbox FREE_X) n))
(unbox FREE_X)))
(define (make-counter x)
(make-closure lifted-anonymous-function-1 (box x)))
Applying the closure would then call the reference to the function, pass the given and the boxed parameters from the closure.
On Mon, Mar 15, 2010 at 4:35 PM,
<mjd-phillylambda@plover.com> wrote:
> So I'd like to test my understanding with this group...
>
> int x;
>
> void foo(int x);
> void baz(void (*F)(void));
>
> void callback(void)
> {
> foo(x);
> }
>
> void bar(void)
> {
> baz(&callback);
> }
>
> C doesn't "support" closures, but the code above has one.
Only in a trivial sense. C doesn't support nested function scopes, so
in C one can't construct an interesting example of closures or of
their failure in C. I wanted to construct a C example for you, but
none of my constructions made sense as C.
To not-answer your question:
> The above is a closure, if you could do such a thing in C.
I think the only thing you could usefully do with this question is to
unask it. Trying to understand closures by studying the non-behavior
of the nonexistent closures of C is pointless.
If you like Perl, I suggest that you read chapter 3 of "Higher-Order
Perl", available for free download at:
http://hop.perl.plover.com/
I think the explanation there is both clear and rigorous.
If you like some other language that has closures, I suggest you ask
for an example in that language.
If you only like C, I suggest you either forget about closures
(because C doesn't have them) or learn to like some language that does
have them.
> The word closure
> is used here because we create an instance of foo() over which x is "closed".
>
> Do I have it right so far?
No:
* You have not created an instance of foo, or indeed of anything.
Functions always have static duration in C. Instances of functions
are created at compile time, and not afterward.
* The jargon is that the function is closed over the variable, not the
other way around.
> I would appreciate it if someone could propose a few problems which
> are trivial to solve with closures, but difficult without.
a. Implement an object-oriented programming system. Closures make
adequate objects. This is discussed in detail in _Structure and
interpretation of Computer Programs_, should you want to know more.
b. Almost anything in _Higher-Order Perl_; see the table of contents.