Dereferencing std::vector passed by reference

Question

I have a quick (and possibly stupid) question, but I couldn't find an answer online:

When I pass a struct by reference, I can nicely and cleanly access its fields via MyArray->Field1, MyArray->Field2 etc.

Is there a similar way to do this for the elements of a std::vector? The only way to do this I know, is something like MyVector[0][0], MyVector[0][1] etc. or maybe do something like std::vector<double> MyVector2 = *MyVector; and then MyVector2[0, 1, 2...] = 1.0...

Edit: Wow, was Stackoverflow quick closing my question...

I think, I got passing by reference and passing by pointer confused. I'm not sure, what the better solution in my case is: Basically I have a class where I set a std::vector* to an "external" std::vector. It uses this vector a lot of times, so I don't want to copy it all the time for performance reasons (and also the class changes it). Passing it by pointer, as mentioned above, I have to write something like MyVector[0][0] = 3; and while this works perfectly, it just doesn't look very nice.

The better way of writing this, according to wohlstads answer (thank you for the good explanation) would be (*MyVector)[0] = 3;.

Answer 1

Passing a std::vector by refernce allows you to access its public members, e.g.:

void f(std::vector<int> & v)
{
    v.resize(1);
    v[0] = 3;
}

However, the syntax you used your question actually looks like passing by pointer. You can do that as well with a std::vector but in C++ we usually prefer to pass by reference. If you have a good reason to pass by pointer instead of a reference, you can do it like that:

void f(std::vector<int> * pV)
{
    pV->resize(1);
    (*pV)[0] = 3;
}

The brackets around *pV are necessary because operator [] has precedence over the dereferencing operator (*); without those, *pV[0] will be interpreted as *(pV[0]).