c++ - Class member qualified name lookup -

c++ - Class member qualified name lookup -

consider next code snippet:

class { int b[a::a]; //1, error void foo(){ int b = a::a; } //2, ok static const int = 5; }

clause 3.4.3.1/1 (qualified name lookup, class members) said:

if nested-name-specifier of qualified-id nominates class, name specified after nested-name-specifier looked in scope of class (10.2)

this implies name a after nested-name-specifier both in //1 , in //2 looked in class scope.

clause 10.2 (member name lookup) said:

10.2/2

the next steps define result of name lookup fellow member name f in class scope c.

10.2/3

the lookup set f in c, called s(f, c)...

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.

the next unclear me:

from quotes cited implies both //1 , //2 same fellow member lookup rules shall applied. but different. why reasoning wrong?

note: know unqualified name lookup rules class scope. , understood behavior in next code snippet:

class { int b[a]; //error void foo(){ int b = a; } //ok static const int = 5; }

it because behavior described in sections 3.4.1/7 , 3.4.1/8 (unqualified name lookup).

the error because when int b[a::a]; beingness processed, a not yet have symbol a. @ point of compilation, a still incomplete because have not reached closing } of class definition yet. compiler doesn't "look ahead" see if future lines of source code contain definition of a.

you can see reversing order of lines:

class { static const int = 5; int b[a::a]; // ok };

the function definition not have same problem because inline function bodies not compiled until after compilation of class definition. (sorry, don't have standard references handy this)

c++ class language-lawyer