On May 3, 2022, at 7:27 PM, Ian Lance Taylor <i...@golang.org> wrote:
>
> Does a program like this print true or false?
>
> func F() func() int { return func() int { return 0 } }
> func G() { fmt.Println(F() == F()) }
>
> What about a program like this:
>
> func H(i int) func() *int { return func() *int { return &i } }
> func J() { fmt.Println(H(0) == H(1)) }
Note that in Scheme eq? works for functions as one would expect.
> (define (f (lambda (x) x))
> (eq? f f)) => #t
> (define g f)
> (eq? f g) => #t
But
> (eq? f (lambda (x) x)) => #f
> (define g (lambda () (lambda (x) x)))
> (eq? (g) (g)) => #f
One can make the case that each closure would be a fresh instance.
This is more clear with a slightly more complex version:
(define (counter m) (let ((n m) (lambda () (set! n (+ n 1)) n))
And equal? is unspecified in the Scheme RnRS but would typically implemented
to return #f.
> (equal? (lambda (x) x) (lambda (x) x)) => #f
Technically (lambda (x) x) & (lambda (y) y) behave identically but
proving the more general case of this in even an interpreter would
be hard to impossible.
Go pretty much has the same considerations (but for a more complex
data model).
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