Exception handling of a function in Python

Question

Suppose I have a function definiton:

def test():
    print 'hi'

I get a TypeError whenever I gives an argument.

Now, I want to put the def statement in try. How do I do this?

Answer 1

try: 
    test()
except TypeError:
    print "error"

Answer 2

In [1]: def test():
     ...:     print 'hi'
     ...:

In [2]: try:
     ...:     test(1)
     ...: except:
     ...:     print 'exception'
     ...:
exception

Here is the relevant section in the tutorial

By the way. to fix this error, you should not wrap the function call in a try-except. Instead call it with the right number of arguments!

Answer 3

You said

Now, I want to put the def statement in try. How to do this.

The def statement is correct, it is not raising any exceptions. So putting it in a try won't do anything.

What raises the exception is the actual call to the function. So that should be put in the try instead:

try: 
    test()
except TypeError:
    print "error"

Answer 4

This is valid:

try:
  def test():
    print 'hi'
except:
  print 'error'


test()

Answer 5

If you want to throw the error at call-time, which it sounds like you might want, you could try this aproach:

def test(*args):
    if args:
        raise
    print 'hi'

This will shift the error from the calling location to the function. It accepts any number of parameters via the *args list. Not that I know why you'd want to do that.

Answer 6

A better way to handle a variable number of arguments in Python is as follows:

def foo(*args, **kwargs):
    # args will hold the positional arguments
    print args

    # kwargs will hold the named arguments
    print kwargs


# Now, all of these work
foo(1)
foo(1,2)
foo(1,2,third=3)