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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)

Eddie Kusturica
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    At least in Racket, it definitely does not. Both `cond` and both `else` identifiers are bound to the new definitions, and the expressions throws an exception for trying to apply `3` as a procedure. – Brendan Cannell May 10 '17 at 15:20
  • Where does that work? As far as I know, it shouldn't work in any Scheme implementation. – molbdnilo May 10 '17 at 21:31

2 Answers2

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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))
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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".

https://www.cs.indiana.edu/~dyb/pubs/bc-syntax-case.pdf

soegaard
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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.

Sylwester
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