On Wed, 18 Jul 2018 at 20:19, Gregg Reynolds <d...@mobileink.com> wrote:
>
> On Wed, Jul 18, 2018, 1:55 PM James Reeves <ja...@booleanknot.com> wrote:
>
>>
>> Function expressions don't evaluate to themselves.
>>
>
> To me that means either the definition is wrong or your literal? is
> rigged. Probably the latter; "literal" is meta, it can't be in the language.
>
How can it be rigged? It's a one line function that just returns true if an
expression evaluates to itself.
You can run read-string directly and see that reading a function expression
results in a seq:
user=> (read-string "#(%)")
(fn* [p1__175#] (p1__175#))
user=> (seq? *1)
true
It doesn't become a function until it's evaluated
Look at it this way: the meaning ("value") of "lambda x.x+1" is just the
> function that increments its arg. Of course you have to "evaluate" to know
> that, but you also have to evaluate "2" in the same way to know what it
> means.
>
Clojure (and lisps in general) distinguish between reading data and
evaluating it. The integer expression "2" does not need to be evaluated:
user=> (read-string "2")
2
user=> (int? 2)
true
And evaluating it only returns itself:
user=> (eval 2)
2
user=> (= 2 (eval 2))
true
In Clojure terminology, you don't need to evaluate "2" to know what it
means. However, you do need to evaluate an expression like (fn [x] (+ x
1)). Merely reading it returns a list.
Anyway my point is that clojure, like lisp, is a form of the lambda
> calculus, where everything is a function, even the Nats like 2. To me at
> least that is one of the most important lessons of functional programming.
> Ymmv.
>
It isn't.
Certainly Clojure can trace some of its lineage back to the lambda
calculus, but that's all. Not everything in Clojure is a function, or even
resembles a function.
--
James Reeves
booleanknot.com
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