How do I break out of a try loop in python?

Question

Please refer to the following code:

import sys
def x():
    try:
        y()
    except:
        print("exception caught")
def y():
    sys.exit()
x()

In this instance, the try loop in function x() will be carried over to the function y() , causing the except loop to run due to sys.exit() raising an error. I know, we can change it to raise SystemExit to exit it, but is there a way to break out of the try loop or is there a better way of writing this code?

Thank you for reading and thanks in advance.

Answer 1

您可以编写except Exception ，它会捕获代码中所有常见的except Exception ，但不会捕获SystemExit异常，因为它不是从Exception继承的，而是从BaseException继承的

Answer 2

In general, it is a bad idea to simply use except without catching any errors... So my advice is to go for the other way you mention, like this:

import sys
def x():
    try:
        y()
    except SystemExit:
        print("exception caught")
def y():
    sys.exit()
x()

Answer 3

I think you want to exit from try block without getting caught by except block, for this just

except Exception as e:

instead of

except:

Here is the full code:

import sys
def x():
    try:
        y()
    except as e:
        if e is SystemExit:
            print("exception caught")
def y():
    raise SystemExit
x()