[racket] question about macro failing due to hygiene(?)

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On 13-01-12 21:24, Eli Barzilay wrote:
> Three hours ago, Marijn wrote:
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>> Why doesn't this work (expand: unbound identifier in module in:
>> a): [...]
> 
> Similar to the reason that this:
> 
> (define-syntax make-store (syntax-rules () [(_ _id_ _E_) (let () 
> (define-syntax-rule (make-variable __E__) (lambda () (let ([_id_
> 1]) __E__))) (define store (let ([_id_ 2]) (make-variable _E_))) 
> store)])) ((make-store a a))

Thanks for the simplified example.

> returns 2 and not 1.

And so it does, but even in the simplified example I find it hard to
reason about it and come to a conclusion on whether it should return 1
or 2.

Do you just look at:

         (define-syntax-rule (make-variable __E__)
           (lambda () (let ([_id_ 1]) __E__)))

and say "__E__ and _id_ are not introduced by the same syntax-rule
pattern, so by hygiene _id_ is not free in __E__ and thus (let ((_id_
1)) ...) cannot bind anything in __E__" or how does it work?

Marijn



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