How to pass optional arguments to a method in C++?

To follow the example given here, but to clarify syntax with the use of header files, the function forward declaration contains the optional parameter default value.

myfile.h

void myfunc(int blah, int mode = 0);

myfile.cpp

void myfunc(int blah, int mode) /* mode = 0 */
{
    if (mode == 0)
        do_something();
     else
        do_something_else();
}

With the introduction of std::optional in C++17 you can pass optional arguments:

#include <iostream>
#include <string>
#include <optional>

void myfunc(const std::string& id, const std::optional<std::string>& param = std::nullopt)
{
    std::cout << "id=" << id << ", param=";

    if (param)
        std::cout << *param << std::endl;
    else
        std::cout << "<parameter not set>" << std::endl;
}

int main() 
{
    myfunc("first");
    myfunc("second" , "something");
}

Output:

id=first param=<parameter not set>
id=second param=something

See https://en.cppreference.com/w/cpp/utility/optional

Jus adding to accepted ans of @Pramendra , If you have declaration and definition of function, only in declaration the default param need to be specified