I have a function Scan.MakeLinearGrid() which returns a vector<vector<Vector3> >

I have another function that is made to accept this type of thing:
Scan.setAngleList(vector<vector<Vector3> >);

But if I call it like this:

Scan.setAngleList(Scan.MakeLinearGrid());

it says

error: initial value of reference to non-const must be an lvalue
    Scan.setAngleList(Scan.MakeLinearGrid());

but if i do

vector<vector<Vector3> > temp = Scan.MakeLinearGrid();
Scan.setAngleList(temp);

it works fine.

Can someone explain the difference?

Thanks,
Dave

Recommended Answers

All 4 Replies

Your Vector3 is missing the appropriate copy constructor.

A good rule of thumb is if you use any one of copy constructor, assignment operator, or virtual destructor, then you need all three. (Not always, but usually.)

Hope this helps.

I have neither an assignment operator or virtual destructor, so then why do i need a copy constructor?

Can you give a simple example of each of those?

What does setAngleList do? When you say Scan.setAngleList(Scan.MakeLinearGrid()); it passes a temporary object to the method. If setAngleList doesn't modify the object, you should make it a const reference anyway because that's both safer and more flexible in what you can pass. If setAngleList does modify the object, you shouldn't be passing a temporary at all, this call has a logical error that's luckily being caught by the compiler, and the working example you gave is the way to fix the error.

Thanks Ed. I'm glad you've got such a sharp eye. I didn't even think of that...

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