C++ access class struct member using an array pointer

Question

Sorry if the title doesn't make sense. I don't know how to word it as I'm fairly new to C++. Basically I have this:

sf::VertexArray *vArray;

If I want to access the position inside, I would have to do this:

(*vArray)[0].position = ...;

Is there a way to use the arrow notation instead? Why can't I do this: vArray[0]->position = ...;?

Any help would be appreciated!

EDIT: sf::VertexArray is part of the SFML library: https://github.com/SFML/SFML/blob/master/src/SFML/Graphics/VertexArray.cpp

Since vArray has type sf::VertexArray *, vArray[0] is just built-in pointer arithmetic, and has type sf::VertexArray. That type has no operator->, and so you cannot do ->position on it. Even if it did have sf::Vertex *operator->(), presumably it would be implemented to return a pointer to the first vertex in the VertexArray, so would only have quite limited use. I expect that's why it's not provided. — Steve Jessop
– Steve Jessop, Commented Sep 11, 2016 at 23:09
If you have dozens of lines that use (*vArray) inside a block, you could reseat it as reference and use that reference (be careful not to delete the vArray pointer before the end of the block) vector<int> *pvi=new vector<int>(2,100); vector<int> &vref=(*pvi); cout << vref[0] << endl; cout << vref[1] << endl; — blackpen
– blackpen, Commented Sep 11, 2016 at 23:30

AnT stands with Russia · Accepted Answer · 2016-09-11 23:08:51Z

3

If your original line

(*vArray)[0].position = ...;

properly illustrates the semantics of your data structure, then the ->-based analogue would be

vArray->operator [](0).position = ...;

assuming sf::VertexArray is a class type with overloaded operator []. Obviously this second form is much more convoluted and requires an explicit reference to the operator member function, which is why it is a better idea to use a much more elegant first form.

Alternatively, you can force a -> into this expression as

(&(*vArray)[0])->position = ...;

but that does not make much practical sense.

You can even combine the two

(&vArray->operator [](0))->position = ...;

to arrive at something even more obfuscated and pointless.

Anyway, why do you want to have a -> in this expression? What are you trying to achieve?

edited Sep 11, 2016 at 23:08

answered Sep 11, 2016 at 23:03

AnT stands with Russia

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1 Comment

syy Over a year ago

Thanks for the explanation. I was just trying to set the position. I'm fairly new to C++ so I don't know how most of the syntax works; especially for pointers.

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