c++ - User-declared namespace member -

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there 3.4.1/14:

if variable member of namespace defined outside of scope of namespace name appears in definition of member (after declarator-id) looked if definition of member occurred in namespace.

if name treated definition of member name point of declaration?

and why following example works:

namespace n  {     extern int j; } int = 2; int n::j = i; //n::j=2

int n::j=i actual appears namespace scope. hence declarationint i=2 not visible unqualified name lookup. why declaration found?

your question:

int n::j=i actual appears namespace scope. hence declaration int i=2 not visible unqualified name lookup. why declaration found?

answer:

since i not found in n namespace, looked in global namespace. had i been there in n namespace, have been used initialize n::j.

hope following program clarifies doubt.

#include <iostream>  namespace n  {    extern int j;    extern int k;     int x = 3; }  int x = 2; int y = 10;  int n::j = x; // n::x used initialize n::j int n::k = y; // ::y used initialize n::k  int main() {    std::cout << n::j << std::endl;    std::cout << n::k << std::endl; }

output:

 3 10

update, in response comment op

what standard saying that:

namespace n  {    extern int j; }  int x = 2;  int n::j = x;

is equivalent to:

namespace n  {    extern int j; }  int x = 2;  namespace n  {    int j = x; }

the logic lookup ofx same. if found within same namespace n, used. if x not found in namespace n, searched outward in enclosing namespaces.


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