Question

How do I test my parameter if it will raise an exception without actually raising it, using try and except?

class MyClass:
    def function(parameter):
        pass

parameter is an ambiguous function that may raise 1 or more of any exception, for example:

parameter = pow("5", 5)

A TypeError is raised as soon as the function is called and before the function can execute its statements.

Answer 1

In a comment to another answer you said: “parameter is another function; take for example: parameter = pow("5", 5) which raises a TypeError, but it could be any type of function and any type of exception.”

If you want to catch the exeption inside your function you have to call the paramenter (which I’m assuming is callable) inside that function:

def function(callable, args=()):
    try:
        callable(*args)
    except:
        print('Ops!')

Example:

>>> function(pow, args=("5", 5))
Ops!

This is if you really need to call your “paramenter” inside the function. Otherwise your should manage its behaviour outside, maybe with something like:

>>> try:
...     param = pow('5', 5)
... except:
...     param = 10
... 
>>> param
10
>>> function(param)

In this example, to raise an exception is pow not function, so it’s a good practice to separate the the two different call, and wrap with a try-except statement the code that might fail.