method overloading python

Question

This is similar to method overloading. When print Hello(1, 2, 3) gets executed it returns "a and b" whereas I want it to return "a and b and c". I know I could have said if (a and b and c is None) then it would have worked. But if I had 20 parameter and I have to handle every case it would be just multiple if statements which I don't think should be necessary. Is there a better way I should be doing such a problem?

 def Hello(a=None, b=None, c=None):
    if(a and b):
        return "a and b"
    if(a and c):
       return "a and c"
    if(a and b and c):
       return "a and b and c"

print Hello(1, 2, 3)

If you had 20 parameters you would need to rethink your design. — BrenBarn
– BrenBarn, Commented Jan 12, 2013 at 21:45
If you had 20 parameters, then no one would use this function. Can you put an actual example of what you want to achieve ? A typical use of overloading is for handling different data type, like int and float, but that is hardly necessary in Python. So, what is your actual use case ? — mmgp
– mmgp, Commented Jan 12, 2013 at 21:47
I was just being hypothetical what about six parameters is the best way to go about it with if statments — user1741339
– user1741339, Commented Jan 12, 2013 at 21:48

Ned Batchelder · Accepted Answer · 2013-01-12 22:01:19Z

5

You need to reorder the if statements:

 def Hello(a=None, b=None, c=None):
    if a and b and c:
       return "a and b and c"
    if a and b:
        return "a and b"
    if a and c:
       return "a and c"

When a and b and c evaluates to true, a and b also evaluates to true, so in your code this case gets handled by the first if.

edited Jan 12, 2013 at 22:01

Ned Batchelder

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answered Jan 12, 2013 at 21:43

NPE

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3 Comments

user1741339 Over a year ago

but my aim is to avoid all the unnecessary if loops if thats possible

Martijn Pieters Over a year ago

@user1741339: Then work out a better API.. How often is this going to happen in real-life software, really?

Matthias Over a year ago

@user1741339: if does not start a loop.

Martijn Pieters · Accepted Answer · 2013-01-12 22:07:48Z

4

You need to rethink your program logic, really. Usually, an API will not have all-optional parameters where you need to build a large decision tree from all the inputs.

In other words, in a real program you won't normally encounter a function where the behaviour is completely different for each of the different combinations of inputs. Invariably, the legal combinations is much, much smaller even for a larger number of optional parameters.

For your toy example, you can come up with other ways to create your return value not involving a decision tree:

map responses:

responsemap = {
    (True,  True,  False): "a and b",
    (True,  False, True):  "a and c",
    (True,  True,  True):  "a and b and c"
}

def Hello(a=None, b=None, c=None):
    return responsemap.get((bool(a), bool(b), bool(c)), None)

build a string based on the inputs:

def Hello(a=None, b=None, c=None):
    return ' and '.join([n for n in ('a', 'b', 'c') if locals()[n]])

edited Jan 12, 2013 at 22:07

answered Jan 12, 2013 at 22:00

Martijn Pieters

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1 Comment

user1741339 Over a year ago

yeah your right. I should rethink my method signature which limits the number of inputs but thanks for the different approaches you showed

Paul Roub · Accepted Answer · 2018-07-02 14:48:00Z

0

python does not supports method overloading. We may overload the methods but can only use the latest defined method

def product(a, b):
    return a*b

def product(a,b,c):
    return a*b*c

if you call product(10, 20) will give an error

edited Jul 2, 2018 at 14:48

Paul Roub

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answered Jul 2, 2018 at 14:46

Prithvi Raj Vuppalapati

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Comments

score 0 · Accepted Answer · 2023-01-28 16:33:06Z

0

this is not best but you can use without class(only function)

def my_func(fname=None, lname=None, job=None):
if fname is not None and lname is not None and job is not None:
    print('Hello', fname, lname)
    print('you are', job)
    print("func 1")

elif fname is not None and lname is not None and job is None:
    print('Hello', fname, lname)
    print("func 2")

elif fname is not None and lname is None and job is None:
    print('Hello', fname)
    print("func 3")


  my_func('dev', 'whale', 'programmer')
  my_func('dev', 'whale')
  my_func('dev')

edited Jan 28, 2023 at 16:33

answered Jan 28, 2023 at 16:32

user19921349

Comments

D3nz13 · Accepted Answer · 2023-01-29 00:41:22Z

0

It's not clear what you want to do. The way I've understood it, you want to check which arguments are not None but the number of arguments is not fixed.

You can use *args or **kwargs for that.

def hello(**kwargs):
    return " and ".join((k for k, v in kwargs.items() if v is not None))


>>> hello(a=1, b=2, c=None)  # kwargs = {'a': 1, 'b': 2, 'c': None}
'a and b'
>>> hello(a=1, b=2, c='b')  # kwargs = {'a': 1, 'b': 2, 'c': 'b'}
'a and b and c'

edited Jan 29, 2023 at 0:41

answered Jan 28, 2023 at 16:54

D3nz13

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