[racket-users] Calling into a Chez Scheme library -- imported symbol rewritten in vm-eval?

I have a Chez Scheme library "foo.ss": (library (foo) (export bar) (import (rnrs)) (define (bar) "hello")) >From Chez Scheme: $ chezscheme Chez Scheme Version 9.5.4 Copyright 1984-2020 Cisco Systems, Inc. > (eval '((lambda () (import (foo)) (bar "hello" A Racket program

Re: [racket-users] Is there a way to eval a custom language from inside a module... without instantiating modules multiple times?

> > 2. Just use the same namespace as eval-example.rkt > I had tried this, but it evals not just the example language but Racket+example (i.e. Racket forms also eval, and the example language has to be compatible with Racket). 1. Keep the new namespace, but use namespace-attach-module to

[racket-users] Is there a way to eval a custom language from inside a module... without instantiating modules multiple times?

I'd like to be able to eval a custom language from inside a module... without instantiating modules multiple times. With the help of Matthew Butterick, I've gotten this far: ; runtime.rkt #lang racket (printf "This is runtime.rkt~n") ; example.rkt #lang racket (require "runtime.rkt")

Re: [racket-users] How to eval a custom language from inside a module?

> > (define-runtime-path example "example.rkt") That works! Thanks! -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to

[racket-users] How to eval a custom language from inside a module?

Suppose I've defined a simple language: ; example.rkt #lang racket (provide foo) (define-syntax foo (syntax-rules () ((foo) (printf "foo!~n" Now I want to create an eval-example function which will evaluate forms in my language: > (eval-example '(foo)) foo! If I'm working

[racket-users] What algorithm does Racket use to test cyclical data structures for equality?

The Racket documentation for equal? says equality is recursively defined; if both v1 and v2 contain reference > cycles, they are equal when the infinite unfoldings of the values would be > equal. I didn't quite believe my ears! :-) So I tried it with two pairs (a b...) and four pairs (a b