What does it mean when a parameter is assigned a value in the function prototype?

Question

I've just seen this syntax for a function prototype in C++:

explicit String(unsigned char value, unsigned char base=10);

I haven't seen this before, but am I right in assuming it sets base to 10 regardless of what you call it with?

Answer 1

The default parameter, called base will take whatever value you send it, or the value 10, if you leave it off, e.g. by calling

String(0);

Given that you can call it with just one parameter, since the second can be defaulted, a constructor can be marked as explicit. This means it won't create a temporary from an unsigned char without you noticing, you have to explicitly call the constructor.