Re: overloading show function
Philipp Schneider <philipp.schneider5 <at> gmx.net>
2011-06-30 23:47:42 GMT
On 06/30/2011 11:46 PM, Holger Siegel wrote:
> Am 30.06.2011 um 22:57 schrieb Philipp Schneider:
>> On 06/30/2011 09:49 PM, Holger Siegel wrote:
>>> (...) But that won't work: After you have evaluated an entry of the environment, you store the resulting
value but you throw away its updated environment. That means, you lose the results of all subcomputations
instead of propagating them to all other copies of the environment. Consider the following expression:
>>> let x = big_computation in let y = x in y + x
>>> First, big_computation is bound to the name x, resulting in an environment [("x", big_computation)].
Then a closure consisting of this environment together with the right hand side 'x' is bound to the name y.
Now y+x is evaluated: The closure is entered, and from its environment the content of x - a call to
big_computation - is looked up. Now big_computation is evaluated and the result is bound to x in this
environment. After that, this result is also returned as the value of y. But when returning from the
evaluation of y, the environment with the updated value of x is lost and you have to re-evaluate it in order
to calculate x+y!
>> Hello Holger,
>> Can you give me an example of a lambda term in which this would be an issue?
>> Evaluating the following works just fine in my implementation.
>> interp (App (Lam "x" (Add (Var "x") (Var "x"))) big_computation) 
>> When the first variable "x" is evaluated my interp function returns the
>> value and the updated environment. Then to evaluate the second variable
>> the value is just looked up from this environment.
>> Of course in the following big_computation would be evaluated twice
>> (App (Lam "x" (App (Lam "y" (Add (Var "x") (Var "y"))) big_computation))
>> But i simply don't have a concept like let x=y.