Python 检查函数是否有返回语句 [英] Python Check if function has return statement

问题描述

例如

def f1():
    return 1

def f2():
    return None

def f3():
    print("Hello")

函数 f1() 和 f2() 会返回一些东西，但 f3() 不会.

Functions f1() and f2() returns something but f3() not.

a = f2()
b = f3()

这里 a 等于 b 所以我不能只比较函数的结果来检查一个函数是否有 return.

And here a equals b so I can't just compare the result of functions to check if one has return or not.

推荐答案

我喜欢 st0le 的检查源的想法，但你可以更进一步，将源解析为源树，这样可以消除误报的可能性.

I like st0le's idea of inspecting the source, but you can take it a step further and parse the source into a source tree, which eliminates the possibility of false positives.

import ast
import inspect

def contains_explicit_return(f):
    return any(isinstance(node, ast.Return) for node in ast.walk(ast.parse(inspect.getsource(f))))

def f1():
    return 1

def f2():
    return None

def f3():
    print("return")

for f in (f1, f2, f3):
    print(f, contains_explicit_return(f))

结果:

<function f1 at 0x01D151E0> True
<function f2 at 0x01D15AE0> True
<function f3 at 0x0386E108> False

当然，这仅适用于具有用 Python 编写的源代码的函数，并非所有函数都适用.例如，contains_explicit_return(math.sqrt) 会给你一个 TypeError.

Of course, this only works for functions that have source code written in Python, and not all functions do. For instance, contains_explicit_return(math.sqrt) will give you a TypeError.

此外，这不会告诉您有关函数的任何特定执行是否命中 return 语句的任何信息.考虑函数:

Furthermore, this won't tell you anything about whether any particular execution of a function hit a return statement or not. Consider the functions:

def f():
    if random.choice((True, False)):
        return 1

def g():
    if False:
        return 1

contains_explicit_return 将在这两个上给出 True，尽管 f 在其执行的一半中没有遇到返回，并且 g永远没有遇到返回.

contains_explicit_return will give True on both of these, despite f not encountering a return in half of its executions, and g not encountering a return ever.

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