当函数用作参数时，try-block不起作用

Question

I have a function which I pass to another function:

def func1(a, b):
    return a/b

The second function is meant to handle exceptions similar to the one raised by func1 :

def func2(func):
    try:
        return func
    except ZeroDivisionError as e:
        print("0")

If I pass

func2(func1(1,2))

The output is 0.5 , as expected. However, func2(func1(1,0)) raises a ZeroDivisionError that is not caught by the except statement:

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)
<ipython-input> in <module>
----> 1 func2(func1(1,0))

<ipython-input> in func1(a, b)
      1 def func1(a, b):
----> 2     return a/b

ZeroDivisionError: division by zero

Can anyone clarify why the except block is not entered in this example?

def func3(a,b):
    try:
        a/b
    except ZeroDivisionError as e:
        print("0")

with input func3(1,0) returns 0 , as expected.

Background : I have a code with multiple API calls, which sometimes return a Connection error. I want to put all these calls inside a function which attempts a retry after a few seconds.

Answer 1

If you must put the check in func2 you will need to make sure that the computation is happening inside func2 . In your case, the call is func2(func1(1,0)) . So what is happening under the hood is, when func2 calls func1(1,0) , it is not computing the func1 inside it, instead it is taking the value of func(1,0) , ie it is asking func1 to do its calculation in its own scope, and return only the value. func1 encounters a ZeroDivisionError without entering the Try block of func2 . Your actual function call becomes func2(return_val_of_func1(1,0)) . So you are essentially checking whether the value of func1 throws any ZeroDivisionError , which is not possible. So instead, what you can do is:

def func1(a, b):
    return a/b

def func2(func1,a,b):
    try:
        return func1(a,b)
    except ZeroDivisionError as e:
        print("0")

func2(func1,1,0)

Answer 2

You have written func2 wrong and are calling it wrong.

Try this instead:

def func2(func):
    try:
        return func()
    except ZeroDivisionError as e:
        print("0")

and call it like this:

func2(lambda:func1(1,2))

Update:

Also note that func2 will return the return value of func() if there is no ZeroDivisionError exception, but when there is an exception it returns None .

Answer 3

func2 will never return ZeroDivisionError as this error was never raised in the try block. In the try block, you are just returning the output of func1 which will never throw an error.

You can try:

def func1(a, b):
    try:
        return a/b
    except ZeroDivisionError as e:
        print("0")


def func2(func):
    return func