c++ - Member name lookup rules -

c++ - Member name lookup rules -

the sec. 10.2 describes fellow member name lookup rules:

10.2/3:

the lookup set f in c, called s(f, c), consists of 2 component sets: declaration set, set of members named f; , subobject set, set of subobjects declarations of these members (possibly including using-declarations) found. in declaration set, using-declarations replaced members designate, , type declarations (including injected-class-names) replaced types designate. s(f, c) calculated follows:

10.2/4:

if c contains declaration of name f, declaration set contains every declaration of f declared in c satisfies requirements of language build in lookup occurs.

consider next 2 examples:

class { void foo(){ a::a; } //s(a, a)={ static const int a; } static const int = 5; }

and

class { int b[a::a]; //s(a, a) empty , programme ill-formed static const int = 5; }

what actual s(f, c) calculation rules , why?

for these code snippets

class { void foo(){ a::a; } //s(a, a)={ static const int a; } static const int = 5; }; class { int b[a::a]; //s(a, a) empty , programme ill-formed static const int = 5; };

you should consider name lookup described in section 3.4 name lookup of standard. have nil mutual quotes cited. though can show s(f, c) illustration name a::a in first class definition. s( a, ) cosists 1 declaration static const int = 5

take business relationship in sec class definition name a::a not found because has declared before usage.

the other rule used name lookup in fellow member functions. in first class definition name a::a found.

all described in section 3.4 of standard pointed out.

as phrase cited more appropriate illustration example following

struct { void f( int ); }; struct b : { using f; void f( char ); };

in case if name f searched s( f, b ) contain 2 declarations

using f; // or void f( int );

and

void f( char );

c++ class language-lawyer