CODEWITHSUNDEEP
pythonpython-2.7python-3.x
Eleutetio
Eleutetio

Reputation: 139

How continue execute program after assertion in python?

I am learning Exception in python and i have some doubt:

Can we use any name as error in raise?

like i have read when you use raise you have to define error type so can't i use any stringname as Error? like SkienaError or i have to keep in mind all the error types and have to use only those Error type names ?

    a=int(input())
if a!=10:
    raise SkienaError
else:
    print(a,"pp")

Second doubt is suppose i want user should input int but he input string so an assert pop up but i want program should continue without terminate and again ask for input until user give int type input , I don't want to use while loop here i want to know if it is possible with raise or assert in python ? like:

a=int(input())
assert type(a)==int
print(a,"hello")

So if user give str type input then is it possible program keep giving error and asking new input until input type is int.

Upvotes: 7

Views: 12626

Answers (2)

Oleg Melnikov
Oleg Melnikov

Reputation: 3298

This might also work for your situation. The function Assert() prints a message sTxt in red color, if the Boolean input bCond is False and interpreter continues execution:

RED, END = '\033[91m', '\033[0m'
printRed = lambda sTxt: print(RED + sTxt + END)
Assert = lambda bCond=False, sTxt='': printRed(sTxt) if not bCond else None

Upvotes: 0

Jordan McQueen
Jordan McQueen

Reputation: 787

In order to make your own exception, you'll have to create it.

e.g.

class MyAppLookupError(LookupError):
'''raise this when there's a lookup error for my app'''

To continue execution after a thrown Exception, do it like this:

a = 5
try:
    assert a == 5
except AssertionError as e:
    print(e)

A try block will attempt to execute a block of code. If an exception occurs, it will execute the except block.

Upvotes: 10

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