c++ - Lifetime of rvalue bound to static const reference -

consider this:

std::string foo();  void bar() {          const std::string& r1 = foo();   static const std::string& r2 = foo(); } 

i know lifetime of string resulting first call foo() extended lifetime of r1.

what temporary bound r2, though? live until end of scope or still there when bar() re-entered?

note: not interested whether particular compiler so. (i interested in 1 use, , can test that.) want know standard has on this.

the temporary extended the lifetime of reference.

[c++14: 12.2/5]: the temporary reference bound or temporary complete object of subobject reference bound persists lifetime of reference except:

• a temporary bound reference member in constructor’s ctor-initializer (12.6.2) persists until constructor exits.
• a temporary bound reference parameter in function call (5.2.2) persists until completion of full-expression containing call.
• the lifetime of temporary bound returned value in function return statement (6.6.3) not extended; temporary destroyed @ end of full-expression in return statement.

• a temporary bound reference in new-initializer (5.3.4) persists until completion of full-expression containing new-initializer. [ example:

  struct s { int mi; const std::pair<int,int>& mp; }; s { 1, {2,3} }; s* p = new s{ 1, {2,3} }; // creates dangling reference 

—end example ] [ note: may introduce dangling reference, , implementations encouraged issue warning in such case. —end note ]

^{(in particular, note none of bullet points match scenario.)}

so, in case, it's until program ends.

we can, of course, test pretty trivially:

#include <iostream>  struct tracked {     tracked() { std::cout << "ctor\n"; };     tracked(const tracked&) { std::cout << "copy\n"; };     ~tracked() { std::cout << "dtor\n"; }; };  void foo() {     static const tracked& ref = tracked(); }  int main() {     std::cout << "main()\n";     foo();     std::cout << "bye\n"; } 

result:

main()
ctor
bye
dtor

(live demo)