How Do I Determine Whether to Use '->' versus '.'

Question

I'm following a D2D sample and therein, this code came up:

    if(!(D2D1_WINDOW_STATE_OCCLUDED & m_renderTarget->CheckWindowState()))
    {
        m_renderTarget->BeginDraw();

        Draw();

        if(m_renderTarget->EndDraw() == D2DERR_RECREATE_TARGET)
        {
            m_renderTarget.Reset();
            Invalidate();
        }
    }

I noticed that the m_renderTarget used both the -> (I forget what it's called) and later uses the dot operator. For some reason I thought that use could only use one or the other depending of if it was a reference type or value type. Apparently I thought wrong but can't find any information about when to use each operator (or more to the point, the purpose of each operator.)

Any clarification would be greatly appreciated.

This looks really strange. What's the type of m_renderTarget? — Axel Gneiting, Mar 17 '13 at 01:24
@AxelGneiting it's far from strange, actually it's very common just look at any smart pointer. ;) — syam, Mar 17 '13 at 01:25
I know what a smart pointer is. It just looked like "Reset()" was some API call on the render target, but it's actually a call on the smart pointer. — Axel Gneiting, Mar 17 '13 at 01:38

score 6 · Accepted Answer · answered Mar 17 '13 at 01:26

I thought that use could only use one or the other depending of if it was a reference type or value type

Usually, yes.

The fact that both are used on the same object in your code indicates it is a "smart pointer" i.e. an object type which has an overloaded operator-> that allows it to act like a pointer.

    m_renderTarget->BeginDraw();

This uses the operator-> to access a member of the object it points to. BeginDraw is a member function of the pointed-to object, not of m_renderTarget.

    m_renderTarget.Reset();

This accesses a member of m_renderTarget itself, not the object it points to. Typically a reset() member replaces the pointed-to object with a null pointer.

So in the first case the -> syntax does something with the object it points to, and in the second case the . syntax does something to the object itself.

Ah yes... it finally clicked in my brain what was going on. Thanks. — Nate222, Mar 17 '13 at 05:16

DotNetUser · Answer 2 · 2013-03-17T01:24:46.437

-1

What the arrow operator does is deference a pointer(pointing towards an object) and then use the ‘ . ‘ operator on the leftover expression.

Its same as

    (*m_renderTarget).BeginDraw();

Arrow operator is used to avoid confusion with this

    *(m_renderTarget.BeginDraw());

which of course could be disastrous and could cause program crashes.

This explained with a good example here

edited Mar 17 '13 at 01:24

answered Mar 17 '13 at 01:18

DotNetUser

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This doesn't explain why both `operator.` and `operator->` are used on the same object (without using `operator*`) – Jonathan Wakely Mar 17 '13 at 01:27

How Do I Determine Whether to Use '->' versus '.'

2 Answers2