Converting string into lambda/value

Question

I want the following code to work:

(define fn.str "(lambda (x) (displayln x)")
(define fn.callable (string->lambda fn.str))
; and then 2 next lines should be valid
(fn.callable 123)
(apply fn.callable '(321))

But best I can get is to eval string and get output, instead my target is to get the actual holding value (λ in this case).

In general, this is also desired effect:

(define val.num "500")
(define val.li "'(1 2 3)")
(define val.fn "(λ (a) a)")
(define num (string->value val.num)) ; => num is now holding 500
(define li (string->value val.li)) ; => list '(1 2 3)
(define fn (string->value val.fn)) ; => callable lambda

Maybe I came close enough to solution, but it's got bad smell.

More info:

Those strings came into program from outside (file/user input).
Extracted/parsed values should be binded to something (hash table, for example) for further access.

Maybe related: stackoverflow.com/questions/20349543/…. Although, if you really want to take arbitrary user strings and eval them, you might need eval. When possible it's preferable to make your own #lang in Racket. — Greg Hendershott
– Greg Hendershott, Commented Mar 9, 2015 at 3:55

Óscar López · Accepted Answer · 2015-03-08 21:13:55Z

2

You're right about something - evaluating code from a string is a bad code smell… maybe there's a better way to do what you want to accomplish. Anyway, here's a possible solution:

(define-namespace-anchor a)
(define ns (namespace-anchor->namespace a))

(define (eval-string str)
  (eval (call-with-input-string str read) ns))

It works as requested for the sample inputs:

(define fn.str "(lambda (x) (displayln x))")
(define fn.callable (eval-string fn.str))
(fn.callable 123)
=> 123

(apply fn.callable '(321))
=> 321

(define val.num "500")
(eval-string val.num)
=> 500

(define val.li "'(1 2 3)")
(eval-string val.li)
=> '(1 2 3)

(define val.fn "(λ (a) a)")
(eval-string val.fn)
=> #<procedure>

answered Mar 8, 2015 at 21:13

Óscar López

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2 Comments

Iskander Sharipov Over a year ago

Using this solution this way: (hash-ref! h key (eval-str "(λ (x y) '(y x))")) causes arity mismatch. Manual check and hash-set! is better choice here (maybe it is obvious, but not for me), at least with this approach.

Óscar López Over a year ago

@IskanderSharipov that sounds more like a problem with the way you're actually using the function when you get it from the hash table, not with the approach itself.

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