From: Rozental, Gennadiy (gennadiy.rozental_at_[hidden])
Date: 2002-11-25 14:25:49
> Your are asking why can't the constructor be not explicit, right?
>
> Well, this would allow the 'direct' syntax fn(1,3) that
> Vincent wanted,
> but...
>
> It would entirely break the pointer semantics because the
> following would be
> allowed:
>
> void foo()
> {
> optional<int> opt ;
> opt = 3 ; // This will construct a temporary optional<int>(3)
> if ( opt == 0 ) // tmp optional<int>(0) here.
> }
You may poison undesired operators.
> Even if it would be allowed to have optional<int&>, you would
> still have to
> use it as if it were a pointer:
>
> void foo ( optional<int&> p = optional<int&>() )
> {
> if ( !!p ) // or ( peek(p) ) or ( initialized(p) )
BTW why do you need 3 methods that doing the same?
> {
> int& the_p = *p ;
> the_p = 3 ;
> etc...
> }
> }
Here 2 questions:
1. Let say we have
struct A { void moo() { ... } };
a. void foo( A* a ) { a->moo(); }
b. void foo( A& a ) { a.moo(); } // here we could assume inlining
Does both above versions are equaivalent form assembler stand point?
2. Now let say we use optional:
void foo( optional<A&> a ) { (*a).moo(); }
>From assembler stand point will it be equivalent to 1.a or 1.b?
And here yet another example where optional<T&> would be useful:
Let say I am using ostream wrapper for my printing and want to separate the
case whgen ostream& is not supplied:
class Log
{
....
optional<ostream&> m_output;
{
And use it for example like this:
template<typename T>
Log::operator<<( T const& t )
{
if( !!m_output )
*m_output << t;
}
Would it be equivalent to reference or pointer? Or it the same in both
cases?
Gennadiy.