The *args and **kwargs is a common idiom to allow arbitrary number of arguments to functions as described in the section more on defining functions in the Python documentation.
The *args will give you all function parameters as a tuple:
def foo(*args):
for a in args:
print(a)
foo(1)
# 1
foo(1,2,3)
# 1
# 2
# 3
The **kwargs will give you all
keyword arguments except for those corresponding to a formal parameter as a dictionary.
def bar(**kwargs):
for a in kwargs:
print(a, kwargs[a])
bar(name='one', age=27)
# name one
# age 27
Both idioms can be mixed with normal arguments to allow a set of fixed and some variable arguments:
def foo(kind, *args, **kwargs):
pass
It is also possible to use this the other way around:
def foo(a, b, c):
print(a, b, c)
obj = {'b':10, 'c':'lee'}
foo(100,**obj)
# 100 10 lee
Another usage of the *l idiom is to unpack argument lists when calling a function.
def foo(bar, lee):
print(bar, lee)
l = [1,2]
foo(*l)
# 1 2
In Python 3 it is possible to use *l on the left side of an assignment (Extended Iterable Unpacking), though it gives a list instead of a tuple in this context:
first, *rest = [1,2,3,4]
first, *l, last = [1,2,3,4]
Also Python 3 adds new semantic (refer PEP 3102):
def func(arg1, arg2, arg3, *, kwarg1, kwarg2):
pass
Such function accepts only 3 positional arguments, and everything after * can only be passed as keyword arguments.
Note:
- A Python
dict, semantically used for keyword argument passing, are arbitrarily ordered. However, in Python 3.6, keyword arguments are guaranteed to remember insertion order.
- "The order of elements in
**kwargs now corresponds to the order in which keyword arguments were passed to the function." - What’s New In Python 3.6
- In fact, all dicts in CPython 3.6 will remember insertion order as an implementation detail, this becomes standard in Python 3.7.
Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.
For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
Best Answer
You need to delimit each parameter name with a ":" at the very least. Technically the name is optional, but it is recommended for readability. So you could write:
or what you suggested: