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Re: Unqualified name lookup

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On Sunday, December 4, 2011 2:23:41 PM UTC+4, Francis Glassborow wrote:
> On 03/12/2011 16:24, koszal wrote:
>
> >
>
> > Hello all,
>
> >
>
> > I'm looking for some explanation of the unqualified name lookup
>
> > specified in the spec.
>
> > In particular there is an example in the point 3.4.1$3:
>
> >
>
> > typedef int f;
>
> > namespace N {
>
> >      struct A {
>
> >          friend void f(A&);
>
> >          operator int();
>
> >          void g(A a) {
>
> >              int i = f(a); // f is the typedef, not the friend
>
> >                              // function: equivalent to int(a)
>
> >          }
>
> >      };
>
> > }
>
> >
>
> > and the text below it says:
>
> > [1] "because the expression is not a function call, the argument-
>
> > dependent name lookup (3.4.2) does not apply and the friend function f
>
> > is not found".
>
> >
>
> > Why the expression 'f(a)' is not a function call?  Could somebody
>
> > explain to me how compiler infers here that this is a functional style
>
> > cast and not a function call?
>
> >
>
> > I just can't wrap my head around this - especially if I take into
>
> > account that the note above this example says:
>
> > [2] "For purposes of determining (during parsing) whether an
>
> > expression is a postfix-expression for a function call, the usual name
>
> > lookup rules apply. The rules in 3.4.2 have no effect on the syntactic
>
> > interpretation of an expression."
>
> >
>
> > My reading of this results in the following expectations for this
>
> > particular example:
>
> > 1. var 'i' is initialized by the expression 'f(a)'
>
> > 2. there is an ambiguity since 'f(a)' could be a function style cast
>
> > or a function call
>
> > 3. 'f' should be looked up for what it is and friend function is found
>
> > 4. ambiguity is resolved and this should be treated as a function call
>
> > 5. there is a typing error which should be reported by the user
>
> > Could anybody fix this list for me - most suspected is point 3 but the
>
> > text [1] says that is not a function call so maybe point 2 is also
>
> > wrong?
>
> >
>
> > I'd appreciate any explanations
>
> >
>
> > Best regards
>
> > Andrzej Ostruszka
>
> >
>
> >
>
>
>
> It was decided (in order to preserve some important techniques) that a
>
> friend function would not be visible unless
>
> 1) the function was declared at namespace scope (i.e. the friend
>
> declaration was insufficient)
>
>
>
> or
>
>
>
> 2) the call was a fully qualified call effectively bringing the class
>
> scope into the search list.
>
>
>
> Now in the code above the definition of g() is treated as if it had
>
> been written at namespace scope as:
>
>
>
> void A::g(A a){
>
>    int i = f(a);
>
> }
>
>
>
> The only f visible at this scope is the typedef and there is nothing
>
> in this relocated code to trigger a search in the scope of A.
>
>
>
> I know it seems odd but that is about par for the course when dealing
>
> with name lookup rules :(
>
>
>
>
>
>
>
> --
>
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>
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>
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@Francis , let me not agree with you, concerning the function g
"> Now in the code above the definition of g() is treated as if it had
>
> been written at namespace scope as:
>
>
>
> void A::g(A a){
>
>    int i = f(a);
"

in the following small example there is a difference

case 1
        class X{
        public:
                X& operator% (const X&) {
                        std::cout << "member fun\n";
                        return *this;
                }

                void f()
                {
                        X& a = *this;
                        a % 1; /* error */
                }

        };

        X& operator%( X& a, int)
        {
                std::cout << "global fun\n";
                return a;
        }

         main ()
         {
               X x;
               x.f();
         }
in this case compiler complains , in function f "a%1"'s look-up ends
with member function and compiler detects argument mismatch

while in the following case 2

        class X{
        public:
                X& operator% (const X&) {
                        std::cout << "member fun\n";
                        return *this;
                }

                void f();
        };

        X& operator%( X& a, int)
        {
                std::cout << "global fun\n";
                return a;
        }

        void X::f()
        {
                X& a = *this;
                a % 1;
        }

         main ()
         {
               X x;
               x.f();
         }
compiler matches on the non-member function


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Unqualified name lookup koszal<krasnal1972@gmail.com> - 2011-12-03 08:24 -0800
  Re: Unqualified name lookup Francis Glassborow <francis.glassborow@btinternet.com> - 2011-12-04 02:23 -0800
    Re: Unqualified name lookup gayan.harutyunyan@googlemail.com - 2012-09-26 17:09 -0700

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