dereferencing boost::shared_ptr<>

[Dennis Jones]

Hello,

Given something like:

boost::shared_p tr<T> t( new T() );


What is the best (correct?) way to dereference the pointer? The following
two methods work. Is there a difference?

T &rt1 = *t.get();
T &rt2 = *t;

And what about getting at the raw pointers:

T *pt1 = t.get();
T *pt2 = &*t;

Is there an advantage to one or the other?

Thanks,

- Dennis

[Joe Gottman]

Re: dereferencing boost::shared_p tr&lt;&gt;


"Dennis Jones" <nospam@nospam. com> wrote in message
news:11eg8g8r92 0s9f7@corp.supe rnews.com...[color=blue]
> Hello,
>
> Given something like:
>
> boost::shared_p tr<T> t( new T() );
>
>
> What is the best (correct?) way to dereference the pointer? The following
> two methods work. Is there a difference?
>
> T &rt1 = *t.get();
> T &rt2 = *t;
>[/color]

I prefer the second method, because you get to type four fewer characters
:-) Also, suppose you have a typedef somewhere that looks like
typedef shared_ptr<T> TPtr;

In the second case, you maintain the option of changing this to
typedef T *TPtr;
without having to change any dereferencing code.
[color=blue]
> And what about getting at the raw pointers:
>
> T *pt1 = t.get();
> T *pt2 = &*t;
>[/color]

I definitely prefer the first case here. What is t.get() == 0. Then the
second case invokes undefined behavior by dereferencing a null pointer. For
example, the boost implementation of shared_ptr asserts that get() != 0
inside its implementations of operator*() and operator->().

Joe Gottman

[Dennis Jones]

Re: dereferencing boost::shared_p tr&lt;&gt;


"Joe Gottman" <jgottman@carol ina.rr.com> wrote in message
news:VbWFe.6689 1$Kp2.3647815@t wister.southeas t.rr.com...[color=blue][color=green]
> >
> > What is the best (correct?) way to dereference the pointer? The[/color][/color]
following[color=blue][color=green]
> > two methods work. Is there a difference?
> >
> > T &rt1 = *t.get();
> > T &rt2 = *t;
> >[/color]
>
> I prefer the second method, because you get to type four fewer characters
> :-) Also, suppose you have a typedef somewhere that looks like
> typedef shared_ptr<T> TPtr;[/color]

Thank you. I agree with your assessment. I was just a little surprised
that the second one compiled and worked. I thought (perhaps naively) it
looked like I was dereferencing the shared_ptr instead of the underlying raw
pointer. But now that I look at the header file, I see that operator *() is
defined to return a reference to the pointee.

[color=blue][color=green]
> > And what about getting at the raw pointers:
> >
> > T *pt1 = t.get();
> > T *pt2 = &*t;
> >[/color]
>
> I definitely prefer the first case here. What is t.get() == 0. Then[/color]
the[color=blue]
> second case invokes undefined behavior by dereferencing a null pointer.[/color]
For[color=blue]
> example, the boost implementation of shared_ptr asserts that get() != 0
> inside its implementations of operator*() and operator->().[/color]

I wouldn't have thought of that -- you're absolutely right.

Thanks for your insight,

- Dennis