Vector and []-operator overloading

Question

I have inherited my class from std::vector. Now I want to overload the []-operator.
When I try to assign a new value to my vector, e.g. v[0]=5, I should rec

Answer 1

This particular error is caused because you are not returning an lvalue, generally defined as something that can appear to the left of an assignment, such as v[0] = 5;. You have more problems as pointed out in the other answers but this is the specific issue you face with that error message ^(a).

The correct specification for overloading the index operator is:

int& operator[] (const int nIndex);

You have to return a reference to the item (so it can be modified) if you want to treat it as an lvalue. The following code shows a fix, although obviously all array indexes map to the same value in this simplified case:

#include 
#include 

class Vec : public std::vector {
    public:
        int& operator[] (int);    // <-- note the '&'
    private:
        int xyzzy;
};

int& Vec::operator[] (int idx) {  // <-- note the '&'
    std::cout << "OK\n";
    return xyzzy;
}

int main () {
    Vec v;
    v[0] = 5;
    v[1] = 6;
    std::cout << v[22] << '\n';
    return 0;
}

The output of this is:

OK
OK
OK
6

In reality, you wouldn't map all indexes to the same value, the code above is simply to illustrate the correct function signature. I haven't bothered to give a more complete example since subclassing classes with non-virtual destructors regularly leads to problems in non-trivial code ^(b).

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^(a) It's not usually considered a good idea to subclass std::vector since the destructor isn't virtual, so you can get into trouble when trying to destroy an object polymorphically.

You're probably better off using a has-a relationship (where your class contains a vector) rather than an is-a relationship (where you inherit).

That unfortunately means you may have to create a lot of pass-through methods from your class to the underlying vector (although only the ones you need) but it will solve the problem with the destructor.

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^(b) See ^(a) :-)