如何将惰性变量传递给函数参数，而无需对其进行求值，除非返回 [英] How to pass lazy variables to a functions parameters without them being evaluated unless returned

问题描述

这个问题是针对python的，尽管我不介意用户分享其他语言的经验.

This question is aimed at python, although I don't mind users sharing experience from other languages.

基本上，我的问题是尝试将惰性变量传递给函数. (就我而言，我可能无法控制该功能，因此无法将其更改为将生成器作为输入).

Basically my problem is trying to pass lazy variables to a function. (in my case i may have no control over the function, so can't change it to take a generator as input).

示例(请注意dict.get是一个函数示例，但很可能是foo)

Example (note that dict.get is an example of a function, but it could well be foo)

def calc():
    sleep(10**100)
    return 42

def my_args():
    yield 'meaning'
    yield calc()


#Instead of

meaning_of_life = dict_.get('meaning', calc())

#I would rather

meaning_of_life = dict_.get(my_args)

我不认为有任何一种优雅的解决方案?可能需要复杂的猴子修补程序吗? (如果是这样，那么我不值得尝试).

I don't suppose there are any one line elegant solutions? Possibly complicated monkey patching is required? (if so it's not worth it for me to try).

谢谢.

推荐答案

您可以使用特殊的对象将计算推迟到首次使用该对象之前.看起来惰性代理可以满足您的要求.

You can use a special object to defer computation until the object is first used. It looks a Lazy Proxy will do what you want.

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