如何从辅助函数中提前返回？

Question

所以我认为我很聪明，DRY 从一堆类似的函数中删除了一堆通用代码，并将它们变成了在一个地方定义的辅助函数。 （见GitHub diff）这样他们就可以从一个地方进行修改。 （见another GitHub diff）

原来是这样的

func_A(stuff):
    if stuff == guard_condition:
        return early
    things = boilerplate + stuff
    do A-specific stuff(things)
    return late

func_b(stuff):
    if stuff == guard_condition:
        return early
    things = boilerplate + stuff
    do B-specific stuff(things)
    return late

我把它改成了

_helper(stuff):
    if stuff == guard_condition:
        return early
    things = boilerplate + stuff
    return things

func_A(stuff):
    things = _helper(stuff)
    do A-specific stuff(things)
    return late

func_B(stuff):
    things = _helper(stuff)
    do B-specific stuff(things)
    return late

但后来我尝试了一下，发现由于我将早期返回（“守卫”？）移到了辅助函数中，它们当然不再工作了。现在我可以轻松地向原始函数添加一些代码来处理这些情况，但似乎没有办法做到这一点，除非再次将复杂性移回各个函数并重复。

处理这种情况最优雅的方法是什么？

Answer 1

这有帮助吗？

 def common_stuff(f):
    def checked_for_guards(*args, **kwargs):
        if stuff == guard_condition:
            return early
        things = boilerplate
        else:
            return f(*args, **kwargs)
    return checked_for_guards

@common_stuff
def func_A(stuff):
    do A-specific stuff(things)
    return late

@common_stuff
def func_b(stuff):
    do B-specific stuff(things)
    return late

Answer 2

您可以将a-specific stuff 和b-specific stuff 提取到核心函数，然后传递给您的辅助函数。然后助手会决定是否调用核心函数：

_helper(stuff, _core_func):
    if stuff == guard_condition:
        return early
    things = boilerplate
    return _core_func(things)

_a_core(_things):
    do a-specific stuff
    return late

_b_core(_things):
    do b-specific stuff
    return late

func_A(stuff):
    return _helper(stuff, _a_core)

func_B(stuff):
    return _helper(stuff, _b_core)

---

在了解帮助者的返回值之前的早期答案

我会给_helper一个返回值：

_helper(stuff):
    if guard:
        return False
    boilerplate
    return True

func_a(stuff):
    if _helper():
        do a-specific stuff
    return

func_b(stuff):
    if _helper():
        do b-specific stuff
    return