Temporaries lifetime in N3290 C++ draft

A point from N3290 C++ draft, § 12.2, 5th point, line 10.

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:

A temporary bound to a reference in a new-initializer (5.3.4) persists until the completion of the full-expression containing the new-initializer. [ Example:

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

— end example ] [ Note: This may introduce a dangling reference, and implementations are en开发者_如何学Ccouraged to issue a warning in such a case. — end note ]

This is the added point when compared to C++03. But the example is not understandable for me. Can you please explain this point with any other example?

I know what dangling references and temporary objects are and that std::pair holds two values of possibly different data types.

---

Temporaies in general last only to the end of the expression that they were created in:

#include <complex>


void func()
{
    std::complex<int>   a; // Real variable last until the end of scope.

    a = std::complex<int>(1,2) + std::complex<int>(3,4);
     // ^^^^^^^^^^^^^^^^^^^^^^  Creates a temporary object
     //                         This is destroyed at the end of the expression.
     // Also note the result of the addition creates a new temporary object
     // Then uses the assignment operator to change the variable 'a'
     // Both the above temporaries and the temporary returned by '+'
     // are destroyed at ';'

If you create a temporary object and bind it to a reference. You extend its lifespan to the same lifespan of the reference it is bound too.

    std::complex<int> const& b  = std::complex<int>(5,6);
                      //           ^^^^^^^^^^^^^^^^ Temporary object
                      // ^^^^                       Bound to a reference.
                      //                            Will live as long as b lives 
                      //                            (until end of scope)

The exception to this rule is when the temporary is bound to a reference in a new initializer.

    S* p1 = new S{ 1, {2,3} };
    // This is the new C++11 syntax that does the `equivalent off`:

    S* p2 = new S {1, std::pair<int,int>(2,3) };
                 //   ^^^^^^^^^^^^^^^^^^^^^^^    Temporary object.
                 //                              This lives until the end of the 
                 //                              expression that belongs to the new.
                 //                              ie the temporary will be destroyed
                 //                              when we get to the ';'

But here we are binding the new temporary object to the member

const std::pair<int,int>& mp;

This is a const reference. But the temporary object it is bound to will be destroyed at the ';' in the above expression so mp will be a reference to an object that no longer exists when you try and use it in subsequent expressions.

}