如何跳出父函数？

如果我想打破一个函数，我可以调用return.

如果我在子函数中并且想跳出调用子函数的父函数怎么办？有没有办法做到这一点？

一个最小的例子：

def parent():
    print 'Parent does some work.'
    print 'Parent delegates to child.'
    child()
    print 'This should not execute if the child fails(Exception).' 

def child():
    try:
        print 'The child does some work but fails.'
        raise Exception
    except Exception:
        print 'Can I call something here so the parent ceases work?'
        return
    print "This won't execute if there's an exception."

这就是异常处理的目的：

def child():
    try:
        print 'The child does some work but fails.'
        raise Exception
    except Exception:
        print 'Can I call something here so the parent ceases work?'
        raise Exception  # This is called 'rethrowing' the exception
    print "This won't execute if there's an exception."

那么父函数将不会捕获异常，并且它将继续在堆栈中向上查找，直到找到捕获异常的人。

如果你想重新抛出相同的异常，你可以使用raise:

def child():
    try:
        print 'The child does some work but fails.'
        raise Exception
    except Exception:
        print 'Can I call something here so the parent ceases work?'
        raise  # You don't have to specify the exception again
    print "This won't execute if there's an exception."

或者，您可以将Exception通过说类似的话来表达更具体的内容raise ASpecificException.