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.
The bit shifting operators do exactly what their name implies. They shift bits. Here's a brief (or not-so-brief) introduction to the different shift operators.
The Operators
>>
is the arithmetic (or signed) right shift operator.
>>>
is the logical (or unsigned) right shift operator.
<<
is the left shift operator, and meets the needs of both logical and arithmetic shifts.
All of these operators can be applied to integer values (int
, long
, possibly short
and byte
or char
). In some languages, applying the shift operators to any datatype smaller than int
automatically resizes the operand to be an int
.
Note that <<<
is not an operator, because it would be redundant.
Also note that C and C++ do not distinguish between the right shift operators. They provide only the >>
operator, and the right-shifting behavior is implementation defined for signed types. The rest of the answer uses the C# / Java operators.
(In all mainstream C and C++ implementations including GCC and Clang/LLVM, >>
on signed types is arithmetic. Some code assumes this, but it isn't something the standard guarantees. It's not undefined, though; the standard requires implementations to define it one way or another. However, left shifts of negative signed numbers is undefined behaviour (signed integer overflow). So unless you need arithmetic right shift, it's usually a good idea to do your bit-shifting with unsigned types.)
Left shift (<<)
Integers are stored, in memory, as a series of bits. For example, the number 6 stored as a 32-bit int
would be:
00000000 00000000 00000000 00000110
Shifting this bit pattern to the left one position (6 << 1
) would result in the number 12:
00000000 00000000 00000000 00001100
As you can see, the digits have shifted to the left by one position, and the last digit on the right is filled with a zero. You might also note that shifting left is equivalent to multiplication by powers of 2. So 6 << 1
is equivalent to 6 * 2
, and 6 << 3
is equivalent to 6 * 8
. A good optimizing compiler will replace multiplications with shifts when possible.
Non-circular shifting
Please note that these are not circular shifts. Shifting this value to the left by one position (3,758,096,384 << 1
):
11100000 00000000 00000000 00000000
results in 3,221,225,472:
11000000 00000000 00000000 00000000
The digit that gets shifted "off the end" is lost. It does not wrap around.
Logical right shift (>>>)
A logical right shift is the converse to the left shift. Rather than moving bits to the left, they simply move to the right. For example, shifting the number 12:
00000000 00000000 00000000 00001100
to the right by one position (12 >>> 1
) will get back our original 6:
00000000 00000000 00000000 00000110
So we see that shifting to the right is equivalent to division by powers of 2.
Lost bits are gone
However, a shift cannot reclaim "lost" bits. For example, if we shift this pattern:
00111000 00000000 00000000 00000110
to the left 4 positions (939,524,102 << 4
), we get 2,147,483,744:
10000000 00000000 00000000 01100000
and then shifting back ((939,524,102 << 4) >>> 4
) we get 134,217,734:
00001000 00000000 00000000 00000110
We cannot get back our original value once we have lost bits.
Arithmetic right shift (>>)
The arithmetic right shift is exactly like the logical right shift, except instead of padding with zero, it pads with the most significant bit. This is because the most significant bit is the sign bit, or the bit that distinguishes positive and negative numbers. By padding with the most significant bit, the arithmetic right shift is sign-preserving.
For example, if we interpret this bit pattern as a negative number:
10000000 00000000 00000000 01100000
we have the number -2,147,483,552. Shifting this to the right 4 positions with the arithmetic shift (-2,147,483,552 >> 4) would give us:
11111000 00000000 00000000 00000110
or the number -134,217,722.
So we see that we have preserved the sign of our negative numbers by using the arithmetic right shift, rather than the logical right shift. And once again, we see that we are performing division by powers of 2.
Best Answer
::
is basically a namespace resolution operator. It allows you to access items in modules, or class-level items in classes. For example, say you had this setup:You could access
CONSTANT
from outside the module asSomeModule::InnerModule::MyClass::CONSTANT
.It doesn't affect instance methods defined on a class, since you access those with a different syntax (the dot
.
).Relevant note: If you want to go back to the top-level namespace, do this: ::SomeModule – Benjamin Oakes