Question

我知道您可以在函数中使用return退出函数，

def function():
    return

但是您可以从子功能中退出父功能吗？

示例：

def function()
    print("This is the parent function")

    def exit_both():
        print("This is the child function")
        # Somehow exit this function (exit_both) and exit the parent function (function)

    exit_both()
    print("This shouldn't print")

function()
print("This should still be able to print")

按照this answer的建议，我尝试举起Exception，但这只是退出了整个程序。

Answer 1

您可以从exit_both引发异常，然后在调用function的地方捕获该异常，以防止程序退出。我在这里使用自定义异常，因为我不知道合适的内置异常，因此应避免捕获Exception本身。

class MyException(Exception):
    pass

def function():
    print("This is the parent function")

    def exit_both():
        print("This is the child function")
        raise MyException()

    exit_both()
    print("This shouldn't print")

try:
    function()
except MyException:
    # Exited from child function
    pass
print("This should still be able to print")

输出：

This is the parent function
This is the child function
This should still be able to print

Answer 2

一个解决方案可能是这样：

returnflag = False
def function():
    global returnflag
    print("This is the parent function")

    def exit_both():
        global returnflag
        print("This is the child function")
        returnflag = True
        return

    exit_both()
    if returnflag == True:
        return
    print("This shouldn't print")

function()
print("This should still be able to print")

或者，如果您不喜欢使用全局变量，可以尝试以下方法：

def function():
    returnflag = False
    # or you can use function.returnflag = False -> the result is the same
    print("This is the parent function")

    def exit_both():
        print("This is the child function")
        function.returnflag = True
        return

    exit_both()
    if function.returnflag == True:
        return
    print("This shouldn't print")

function()
print("This should still be able to print")

从子函数的父函数返回

2 个答案: