Question

Coming from C, this example in Chapter 5 of Accelerated C++ caught my attention:

vector<Student_info> extract_fails(vector<Student_info>& students) {

    vector<Student_info> pass, fail;

    for (vector<Student_info>::size_type i = 0; i != students.size(); ++i)
        if (fgrade(students[i]))
            fail.push_back(students[i]);
        else
            pass.push_back(students[i]);

    students = pass;
    return fail;
}

Since fail is returned, I know it’s not a problem for it to be a local variable. But why is pass able to make it out of the local scope?

Answer 1

You probably believe that the reference students is being “re-referenced” to the local variable pass. This is not the case. Instead, operator= is being called on the existing object passed in through students, and the data from pass is copied to it.

As you’re coming from C, you’re probably more familiar with pointers than with references. Let’s assume that students was declared as a pointer, i.e.

vector<Student_info>* students

Your question implies that you believe the code you quoted is equivalent to the following

students = &pass;

when in fact it is equivalent to this

*students = pass;

The first of these two operations cannot, in fact, be carried out on a reference — references cannot be “re-referenced”.