Re: is this a reference to a tem****ary or not?

Joshua Lehrer wrote:

> Does the following invoke undefined behavior? 

Yes.

> The question is whether 
> or not a tem****ary is created in the call to "cast".  If so, then cast
> is returning a reference to a tem****ary, and thus "Const" is returning
> a reference to a tem****ary which then falls out of scope.

Yes, a tem****ary is created.

> 
> The attempt is to convert "unsigned int *" to "const unsigned int *"
> via a function call.  Please, I know, use const_cast, etc...  This is
> a very boiled down example of a much larger problem.
> 
> 
> #include <iostream>
> 
> typedef const unsigned int * const ret_type;
> 
> static inline ret_type& cast(ret_type &param) { return param; }
> 
> inline ret_type & Const(unsigned int * const & param) { return
> cast(param); }
> 
>   inline void print(const unsigned int * const start,
>                        const unsigned int * const finish) {
>    std::cout << start << ' ' << finish << std::endl;
> }
> 
> int main()
> {
>    unsigned int arr[2] = {42, 43};
>    print(&arr[0], &arr[1]);
>    print(Const(&arr[0]), Const(&arr[1]));

The address-of operator (&) yields an r-value of the non-class
type 'pointer to unsigned int'.
When binding a non-class type r-value to a reference, the standard
states quite clearly in section [decl.init.ref]/5, last bullet that a
tem****ary gets created.

> }
> 
> 
> thanks,
> 
> joshua
> 
Bart v Ingen Schenau
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