How can I achieve below in Scheme REPL? Create a variable name from a string.
=>(define (string->variable-name "foo") 12)
=>foo
12
=>(+ foo 8)
20
In Common Lisp, this should be
=> (set (intern "ANY-TEXT") 5)
=> ANY-TEXT
5
How do I build a #procedure like "string->variable-name" (and "variable-name->string") ?
Thanks a lot.
If what you're passing to string->variable-name is always a string literal (i.e., not a variable that contains a string), you can do that using a syntax-case macro that transforms the string literal to an identifier:
(define-syntax string->variable-name
(lambda (stx)
(syntax-case stx ()
((_ str)
(string? (syntax->datum #'str))
(datum->syntax #'str (string->symbol (syntax->datum #'str)))))))
and conversely, a variable-name->string macro could look like this:
(define-syntax variable-name->string
(lambda (stx)
(syntax-case stx ()
((_ id)
(identifier? #'id)
(datum->syntax #'id (symbol->string (syntax->datum #'id)))))))
However, remember: this will only work if you are working with string (in case of string->variable-name) or identifier (in case of variable-name->string) literals.
If, on the other hand, you want the ability to reflect on names in your current Scheme environment, this is not supported by standard Scheme. Some implementations, like Guile or Racket, do have this capability.
Here's a Guile example:
> (module-define! (current-module) 'foo 12)
> foo
12
> (define varname 'bar)
> (module-define! (current-module) varname 42)
> bar
42
and a Racket example:
> (namespace-set-variable-value! 'foo 12)
> foo
12
> (define varname 'bar)
> (namespace-set-variable-value! varname 42)
> bar
42
string->syntax-name? As I mentioned in my answer, it only works when using string literals, e.g., (string->syntax-name "foo"). You cannot use (let ((foo "bar")) (string->syntax-name foo)); I have a syntax guard in place to block that, and the error message suggests to me that you were indeed attempting such a thing and my guard caught it.string->variable-name as a regular #procedure, which also (define foo "bar")(string->variable-name foo) should work, and bar will also be a variable now. I want to use it as a procedure, but thank you :-)eval? I keep getting the feeling that we're dealing with an XY problem here.Forgetting the syntax for a moment:
(define string->variable-name string->symbol)
(define name->value-mapping '())
(define (name-set! name value)
(set! name->value-mapping
(cons (cons name value)
name->value-mapping))
value)
(define (name-get name)
(cond ((assoc name name->value-mapping) => cdr)
(else 'unbound)))
Sure you can't do (+ 5 <my-named-value>) without some other help...
