What is the class keyword before a function argument?

Question

Why does this code work? See the class keyword in front on the f function argument? What does it change if I add it?

struct A
{
    int i;
};

void f(class A pA) // why 'class' here?
{
    cout << pA.i << endl;
}

int main() 
{
    A obj{7};
    f(obj);
    return 0;
}

Answer 1

If a function or a variable exists in scope with the name identical to the name of a class type, class can be prepended to the name for disambiguation, resulting in an elaborated type specifier.

You are always allowed to use an elaborated type specifier. Its major use case, however, is when you have a function or variable with an identical name.

Example from cppreference.com:

class T {
public:
    class U;
private:
    int U;
};

int main()
{
    int T;
    T t; // error: the local variable T is found
    class T t; // OK: finds ::T, the local variable T is ignored
    T::U* u; // error: lookup of T::U finds the private data member
    class T::U* u; // OK: the data member is ignored
}