4 ms·
Lisp runtime implicitly runs each top-level expression as statement. When you run (def x y) it's not only true that this expression returns 'x, it's also true t
by gnulinux 6y ago
Lisp runtime implicitly runs each top-level expression as statement. When you run (def x y) it's not only true that this expression returns 'x, it's also true that it alters the static state of the program by adding 'x to the scope.
To make a purely statementless language, I think you need to make everything happen in a single monadic computation. E.g. in pseudo-Haskell-ish:
main : Scope -> StaticIO Scope
main scope0 = do
scope1, x' <- def scope0 "x" Int
scope2, xv <- dec scope1 "x" 3
But the main part is still statement.
- tsimionescu 6y agoAt least in all of the discussions I have seen, the difference between statements and expressions is whether they return a value, not whether they have side-effects. So (def x b) is an expression, because it returns a value. {} in C is a statement, even though it has no side effects. Now, there are some statements in CL as far as I remember - I believe (declaim x integer) is a kind of statement, at least in the sense above. But you can write a CL program that doesn't use a single statement.
- fiddlerwoaroof 6y agoThe return value of some forms is implementation dependent, but all forms return a value in CL: CL-USER(1): (defparameter *foo* (declaim (special x))) *FOO* CL-USER(2): *foo* (X) CL-USER(3): (defparameter *bar* (declaim (type integer x))) *BAR* CL-USER(4): *bar* (X) In other cases, e.g. `(values)` which is an expression that returns no values, the value is nil