I have this task where i'm working with a metacircular evaluator, and i define a new cond
like this:
(define cond 3)
As well as else
:
(define (else x) (/ x 2)
My question is why does this (below) actually work?
(cond ((= cond 2) 0)
(else (else 4)))
How does Scheme know which cond
is my defined cond and my else
, over the conditional cond
and else
?
(Feel free to edit the title, as i'm not sure how to formulate my question)
In Scheme there are no reserved identifiers. In many languages there is a list of reserved identifiers (keywords) that can't be used as names of variables.
In Scheme you can for example do this:
> (let ((cond +))
(cond 1 2))
3
What sets Scheme apart from most languages is that programs are macro expanded.
Running a Scheme program:
read -> macro expansion -> compilation -> execution
It is non-trivial to explain how the macro expansion algorithm works. I can recommend the chapter "Syntactic Abstraction: The syntax-case expander" by R. Kent Dybvig in the book "Beautiful Code".
It depends on how you have implemented cond
in the metacircular evaluator. Usually it checks some operators for symbols like quote
and cond
and then do someething special. Thus cond
in operator position will be expanded as cond
while cond
in other circumstances would be evaluated as if it was a variable.
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