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

A declaration like

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

clearly shows in the declaration, that you want to fulfill a requirement with your function - e.g. override a specific function in the base class or interface, which e.g. could be declared there as

virtual void foo1(int something, int another, int andAnother) = 0;

BUT you don't intend to use the parameters which are handed over to you.

Another example would be if you want to hand over the function to e.g. another function which expects a function pointer to a void function with three int parameters.

void giveMeFoo( void (*fn)(int, int, int) ) { ... }

Additionally, higher warning levels issue a warning, if parameters are declared, which are not evaluated in the function body. You can avoid that by leaving the parameter names away.

The parameters without names are then indeed not accessible in the function body anymore - on purpose. user4581301 has nicely described, why.

Declaring a standalone function without parameter names as in your example is allowed because of the usages described above, but it obviously makes no sense in most cases. An example where it does make sense is in the comment section. Another example of a standalone function without parameter names could be, if your'e writing a library and want to either maintain backward compatibility (your library function does not need the parameter anymore, but you don't want to break the public header declaration) or you want to reserve a parameter for future use.