making an argument optional?

[4Gb4jE8b]

I have a function

void type_erase(std::string content, int placement, int backspace, double time_delay)

but i want to include an optional argument like so

void type_erase(std::string content, int placement, int backspace, double time_delay, OPTION)

can i do so without copy/pasting the entire function? Lets say that OPTION is specifically an int. and if so, generally speaking, how would i do so?

[Bazzy]

Here is an example void foo ( int a, int b = 0 )
There 'a' is a normal parameter, 'b' is optional, if no value is passed for 'b', 0 will be used for it
After you declare an optional parameter, all following parameters must be optional too
If you have the prototype and implementation of the function in two separate places, you must specify the default value only in the prototype
eg

void foo ( int a, int b = 0 ); // or void foo ( int , int = 0 )

void foo ( int a, int b )
{
     //...
}

[wolfgang]

Like bazzy said. Just use default parameters:

#include <iostream>

int count_to(int top, int steps = 1)
{
   for (int i = 0; i < top; i+=steps)
      std::cout << i << std::endl;

   return 0;
}

int main()
{
   count_to (10);
   count_to(100, 10);
}

[4Gb4jE8b]

THANK YOU! that's exactly what i needed