How to Define a Function with Optional Arguments

How do I define a function with optional arguments?

Try calling it like: obj.some_function( '1', 2, '3', g="foo", h="bar" ). After the required positional arguments, you can specify specific optional arguments by name.

Is there any way to define a Python function with leading optional arguments?

A single function is not allowed to have only leading optional parameters:

8.6. Function definitions

[...] If a parameter has a default value, all following parameters up until the “*” must also have a default value — this is a syntactic restriction that is not expressed by the grammar.

Note this excludes keyword-only parameters, which never receive arguments by position.

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If desired, one can emulate such behaviour by manually implementing the argument to parameter matching. For example, one can dispatch based on arity, or explicitly match variadic arguments.

def leading_default(*args):
    # match arguments to "parameters"
    *_, low, high, size = 0, *args
    print(low, high, size)

leading_default(1, 2)     # 0, 1, 2
leading_default(1, 2, 3)  # 1, 2, 3

A simple form of dispatch achieves function overloading by iterating signatures and calling the first matching one.

import inspect

class MatchOverload:
    """Overload a function via explicitly matching arguments to parameters on call"""
    def __init__(self, base_case=None):
        self.cases = [base_case] if base_case is not None else []

    def overload(self, call):
        self.cases.append(call)
        return self

    def __call__(self, *args, **kwargs):
        failures = []
        for call in self.cases:
            try:
                inspect.signature(call).bind(*args, **kwargs)
            except TypeError as err:
                failures.append(str(err))
            else:
                return call(*args, **kwargs)
        raise TypeError(', '.join(failures))

@MatchOverload
def func(high, size):
    print('two', 0, high, size)

@func.overload
def func(low, high, size):
    print('three', low, high, size)

func(1, 2, size=3)    # three 1 2 3
func(1, 2)            # two 0 1 2
func(1, 2, 3, low=4)  # TypeError: too many positional arguments, multiple values for argument 'low'

Correct way to specifiy optional arguments in R functions

You could also use missing() to test whether or not the argument y was supplied:

fooBar <- function(x,y){
    if(missing(y)) {
        x
    } else {
        x + y
    }
}

fooBar(3,1.5)
# [1] 4.5
fooBar(3)
# [1] 3

Adding optional arguments to a function

Optional[list] means that the argument can either be a list or None. You are still required to pass it from the caller.

If you want an argument that can be omitted you should use a default value, for example:

def function(
    data1: list,
    data2: list,
    opt1: Optional[list] = None,
)

assuming that None means an omitted argument in your function logic.

How can I make a function with optional arguments in C?

The open function is declared as a variadic function. It will look something like this:

#include <stdarg.h>

int open(char const * filename, int flags, ...)
{
    va_list ap;
    va_start(ap, flags);

    if (flags & O_CREAT)
    {
        int mode = va_arg(ap, int);
        // ...
    }

    // ...

    va_end(ap);
}

The further arguments are not consumed unless you have indicated that they do in fact exist.

The same construction is used for printf.

The manual doesn't always make this explicit, since the only possible two signatures are (char const *, int) and (char const *, int, int), so there's little point in revealing that you the function actually accepts variable arguments. (You can test this by trying to compile something like open("", 1, 2, 3, 4, 5, 6).)

How to write function for optional parameters in postgresql?

You can define optional parameters by supplying a default value.

create function foo(p_one integer default null, 
                    p_two integer default 42, 
                    p_three varchar default 'foo')
  returns text
as
$$
begin
    return format('p_one=%s, p_two=%s, p_three=%s', p_one, p_two, p_three);
end;
$$
language plpgsql;

You can "leave out" parameters from the end, so foo(), foo(1) or foo(1,2) are valid. If you want to only supply a parameter that is not the first you have to use the syntax that specifies the parameter names.

select foo(); 

returns: p_one=, p_two=42, p_three=foo

select foo(1); 

returns: p_one=1, p_two=42, p_three=foo

select foo(p_three => 'bar')

returns: p_one=, p_two=42, p_three=bar

python how to define function with optional parameters by square brackets?

1. "if we can define optional parameters using this way(no at present)"

   The square bracket notation not python syntax, it is Backus-Naur form - it is a documentation standard only.

2. A module-level function is a function defined in a module (including __main__) - this is in contrast to a function defined within a class (a method).

Optional arguments in nested functions in Python

Maybe try the code snippet below

def foo(a=1,b=2,c=3):
    d = a + b + c
    return d

def bar(variable: str):
    z = int(input())
    e = foo(**{variable: z})
    return e

# variable name should be defined as string
print(bar("a"))

The ** parses all arbitrary arguments on dict, in this case is a. Be careful tho, as if passing wrong variable name (differ from a, b or c) will result raising error.

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