Why is a function without argument identifiers valid in C++?

Question

Given a function in C++ with arguments that are only types and have no identifiers,

 void foo1(int, int, int){cout << \"called foo1\";}

I

Answer 1

Yes. It is legal in C++.

C++11 n3337 standard 8.4.1(p6) Function definitions:

Note: Unused parameters need not be named. For example,

void print(int a, int) {
    std::printf("a = %d\n", a);
}

C++14 standard:

[ 8.3.5.11] An identifier can optionally be provided as a parameter name; if present in a function definition , it names a parameter (sometimes called “formal argument”). [Note: In particular, parameter names are also optional in function definitions and names used for a parameter in different declarations and the definition of a function need not be the same.]