[Python-ideas] Another way to avoid clumsy lambdas, while adding new functionality
Andrew Barnert
abarnert at yahoo.com
Wed Mar 5 02:16:38 CET 2014
On Mar 4, 2014, at 17:05, João Bernardo <jbvsmo at gmail.com> wrote:
>> * assign a bill to `a` so that `a` will evaluate to the value of the name `foo` any time that `a` is evaluated, in the scope of that evaluation
>>
>> a = $foo
>>
>> * as above, but always plus one
>>
>> a = $foo + 1
>>
>> * make `a` a bill that evaluates to the value of the name `foo` at the time that `a` is evaluated, in that scope, plus the value of `bar` **at the time and in the scope of the assignment to `a`**
>>
>> a = $foo + bar
>
>
> You can do all that programmatically. No need for magic operators.
> Check this module as a proof of concept: https://github.com/jbvsmo/funcbuilder
>
> foo = FuncBuilder()
> a = foo + 1
> # Then evaluate a with a function call
>
> For it to do exactly what you want, you just need override the evaluation function (Funcbuilder.__call__) to load names from current frame locals or something.
> BTW, I don't recommend using that on real code because it is extremely experimental (unless you're a hipster... In this case, carry on).
I posted something similar in the first of the many improving-lambda threads this month, but it had a lot of little problems and one big one.
The big one is that not every expression can be caught by operator overloading (foo[i] is fine, but lst[foo] isn't). And call expressions are one of the most important kinds of expression, but you can't catch that--or, rather, if you _do_ catch it, then you have no way to call the resulting object.
The same idea in C++ doesn't have that last problem because C++ is statically typed, and you can use an implicit cast from autolambda object to function object to get from an object whose call operator builds a call expression to one whose call operator evaluates the expression. But that doesn't work in Python.
I wrote a blog post that gets into this further, but I can't find the link from my phone. I included it in an earlier message in the thread.
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