Passing value to const reference parameter?

Question

I haven't been able to find a similar question on here, or through Google, maybe because i don't know if I'm asking the right question, so not sure if the title is right. But I'm working with code that looks like this:

Foo * foo = new Foo();
add(*foo); //Couldnt find any similar questions on this syntax?

void add(const Foo & foo){

//What exactly is foo here? How can i manipulate its members, foo. or foo-> ?

} 

Answer 1

The variable foo inside the add method is a constant reference. It behaves like a pointer in the sense that it does not copy the object: you interact with it as if it were the object itself. Unlike a pointer, a reference cannot be NULL, and it cannot be reassigned. You interact with it using a dot . operator instead of the -> operator.

void add(const Foo & foo){
    cout << foo.first_member << endl;
    foo.const_member_function();
}

Since foo is a constant reference, you can read foo's members, but you cannot assign them. Also, any calls of member functions on the foo reference are limited to functions explicitly marked const.

Calling add with a dereferenced foo is OK, as long as foo is not NULL. Making a reference by dereferencing a NULL pointer is undefined behavior.

Answer 2

'foo' is a constant reference. You can access its members via foo.fred, foo.method(), etc., but you can't change any data members or call any non-const methods.