[boost] Re: Interest in smart pointer with value semantics?

From: David B. Held (dheld_at_[hidden])
Date: 2004-01-03 22:45:04

"Thorsten Ottosen" <nesotto_at_[hidden]> wrote in message
news:bt7rbb$9dl$1_at_sea.gmane.org...
> "David B. Held" <dheld_at_[hidden]> wrote in message
> news:bt7931$u7g$1_at_sea.gmane.org...
> [...]
> > I think you're misled by the fact that a const member function
> > can only call other const *member functions*.  But there is nothing
> > intrinsically wrong with a const member function calling a non-
> > const free function.
>
> I think there is. It's a giant loophole in the general const-correcness
> of all programs.

Then let me try to convince you otherwise with an example:

void log(std::ostream& s, int x);

class foo
{
    int x_;
public:
    int x(void) const { log(some_log, x_); return x_; }
};

Is this code not const-correct?

> [...]
> 2) in generic code, const T* and const smart_ptr<T> are not equal
> and I will need traits to deal with different syntax
> [...]

As Jeremy points out, they ought not to be equal, either.  To see
why, let's level the playing field:

template <typename T>
struct raw_ptr { typedef T* type; };

Now, how do you spell "const T*" with this type generator?  Why,
you spell it:

raw_ptr<const T>::type

Doesn't that look suspiciously similar to:

smart_ptr<const T>

?

It looks like you are missing the fact that const smart_ptr<T> is
really analogous to T* const.  Now, if you are arguing that T* const
should be the same as T const*, that's an entirely different ball of
wax.  But I'm still not convinced.

Dave

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