Python's Lazy Evaluation on Exception Handling

I tried a simple code like this:

[sourcecode lang="python"]
try:
while True:
print('yipee')
except KeyboardInterrupt:
print('w00t')
[/sourcecode]

It runs successfully on command line interface, and keeps printing "yipee" until the user press Ctrl-C. After that I did a typo, which turns out to mistype the "KeyboardInterrupt" as "KeybaordInterrupt".

[sourcecode lang="python"]
try:
while True:
print('yipee')
except KeybaordInterrupt:
print('w00t')
[/sourcecode]

To my surprise, it still runs well, it keeps printing "yipee" to the screen. It's just that
when I press Ctrl-C, Python threw this error:

[sourcecode lang="shell"]
Traceback (most recent call last):
File '<stdin>', line 3, in <module>
KeyboardInterrupt
[/sourcecode]

During handling of the above exception, another exception occurred:

[sourcecode lang="shell"]
Traceback (most recent call last):
File '<stdin>', line 4, in <module>
NameError: name 'KeybaordInterrupt' is not defined
[/sourcecode]

I looked at the documentation of Python 3.1 which says:
The try statement works as follows.

First, the try clause (the statement(s) between the try and except keywords) is executed.

• If no exception occurs, the except clause is skipped and execution of the try statement is finished.


• If an exception occurs during execution of the try clause, the rest of the clause is skipped. Then if its type matches the exception named after the except keyword, the except clause is executed, and then execution continues after the try statement.


• If an exception occurs which does not match the exception named in the except clause, it is passed on to outer try statements; if no handler is found, it is an unhandled exception and execution stops with a message as shown above.

The exception handling in Python works in a lazy manner, they will only validate the except clauses just before processing the exception!