I tried to evaluate this: (define lambda (lambda (x) x)). MIT Scheme 11.2 gives an error: ;Unbound variable: x. Chez Scheme 9.5 also gives an error: Exception: variable x is not bound. Why is x not bound? I thought that define would evaluate (lambda (x) x) into an anonymous function, and then define lambda to be that anonymous function. Where does x get involved?
I don't get any errors in Racket 7.2 and Guile 3.0.1.
This isn't valid Scheme code:
(define lambda (lambda (x) x))
R7RS seems pretty clear about forbidding this in Section 5.4 about syntax definitions:
However, it is an error for a definition to define an identifier whose binding has to be known in order to determine the meaning of the definition itself, or of any preceding definition that belongs to the same group of internal definitions.
In the posted code the identifer lambda is being redefined, but the binding of lambda itself must be known in order to determine the meaning of the new definition; the above language forbids this.
R6RS has some similar language in Chapter 10 about the expansion process, but the R6RS language is more tightly coupled to the technical details of the expansion process. I'm pretty sure that it applies in the same way to this case, but not 100% sure.
A definition in the sequence of forms must not define any identifier whose binding is used to determine the meaning of the undeferred portions of the definition or any definition that precedes it in the sequence of forms.
OP notes the error message Exception: variable x is not bound and asks: "Where does x get involved?"
Granting that (define lambda (lambda (x) x)) is malformed and thus not valid Scheme, it isn't too meaningful to try to explain such behaviors. Yet it seems that this sort of behavior can be triggered by attempting to redefine other syntactic keywords in similar fashion. Consider:
> (define if (if x y z))
Exception: variable z is not bound
It seems obvious here that z is unbound, so the error doesn't seem unreasonable. But now:
> (define if (if #t 1 2))
Exception: variable if is not bound
Even if is unbound in the right-hand expression of the define form! If we assume that something similar is happening in (define lambda (lambda (x) x)) then lambda is unbound in the right-hand expression, and if that is the case, then (lambda (x) x) is not evaluated as a special form, but as an ordinary procedure call. The order of evaluation for arguments in a procedure call is unspecified, so it is perfectly reasonable that x could be reported as unbound before lambda in this case. We can get rid of the appearances of unbound xs to see if lambda is being bound:
> (define lambda (lambda))
Exception: variable lambda is not bound
> (define lambda lambda)
Exception: variable lambda is not bound
So it seems that the problem here is that attempting to redefine a syntactic keyword using an expression that relies on that keyword for its meaning can cause the binding for the syntactic keyword to become inaccessible.
In any case, take this last bit with a grain of salt because in the end this sort of redefinition is not valid code, and no particular behavior should be expected of it.
(define list (list 1 2))? besides, why would you ever want to write such a code? nobody ever would / should, so if a certain implementation has troubles with it, so what? :) - Will Nesslist, as you say, so maybe it't not that bad. that wasn't the most successful illustration. :) still it brought the point across, hopefully. to do that withlambdathough ..... why?? :) - Will Nessdefineform, but that didn't sit quite right. It seems that the problem was really connected to redefining a keyword in a way that relies on the original binding of that keyword for the meaning of the new definition. - ad absurdum