> I can't produce a more thorough answer now, but in general the whole
> uncovered expressions feature relies on expressions with a source
> indication. For example, this works as you expect:
> ...
> and it should also work for a sandbox that is created from a file.
OK, thanks, that sort of makes sense. But still, if I put the text of the
program in a file and do the following it still produces '()...
(define Ev
(call-with-input-file* "test.rkt"
(λ(inp)
(parameterize ([sandbox-coverage-enabled #t])
(make-evaluator 'lang/htdp-intermediate
inp
)))))
(get-uncovered-expressions Ev)
And so does this:
(define Ev
(parameterize ([sandbox-coverage-enabled #t])
(make-evaluator 'lang/htdp-intermediate
(open-input-string
"(define (f x)
(if (zero? x) \"zero\" \"non-zero\"))
(f 4)")
)))
(get-uncovered-expressions Ev)
--- nadeem
On Jul 21, 2010, at 6:19 PM, Eli Barzilay wrote:
> On Jul 21, Nadeem Abdul Hamid wrote:
>> Sorry if I am missing something obvious, by why does the following
>> produce '() ?
>>
>> (define Ev
>> (parameterize ([sandbox-coverage-enabled #t])
>> (make-evaluator 'lang/htdp-intermediate
>> `(define (f x)
>> (if (zero? x) "zero" "non-zero"))
>> `(f 4)
>> )))
>> (get-uncovered-expressions Ev)
>>
>>
>> I would have expected it to indicate somehow that the "zero" case is
>> not covered?
>
> I can't produce a more thorough answer now, but in general the whole
> uncovered expressions feature relies on expressions with a source
> indication. For example, this works as you expect:
>
> (define Ev
> (parameterize ([sandbox-coverage-enabled #t])
> (make-evaluator 'lang/htdp-intermediate
> "(define (f x)
> (if (zero? x) \"zero\" \"non-zero\"))
> (f 4)"
> )))
> (get-uncovered-expressions Ev)
>
> and it should also work for a sandbox that is created from a file.
>
> --
> ((lambda (x) (x x)) (lambda (x) (x x))) Eli Barzilay:
> http://barzilay.org/ Maze is Life!
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