Aggregate reference member and temporary lifetime

Question

Given this code sample, what are the rules regarding the lifetime of the temporary string being passed to S.

struct S
{
    // [1] S(const std::         


        

Answer 1

The lifetime of temporary objects bound to references is extended, unless there's a specific exception. That is, if there is no such exception, then the lifetime will be extended.

From a fairly recent draft, N4567:

The second context [where the lifetime is extended] 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:

• (5.1) A temporary object bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.
• (5.2) The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not extended; the temporary is destroyed at the end of the full-expression in the return statement.
• (5.3) 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.

The only significant change to C++11 is, as the OP mentioned, that in C++11 there was an additional exception for data members of reference types (from N3337):

• A temporary bound to a reference member in a constructor’s ctor-initializer (12.6.2) persists until the constructor exits.

This was removed in CWG 1696 (post-C++14), and binding temporary objects to reference data members via the mem-initializer is now ill-formed.

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Regarding the examples in the OP:

struct S
{
    const std::string& str_;
};

S a{"foo"}; // direct-initialization

This creates a temporary std::string and initializes the str_ data member with it. The S a{"foo"} uses aggregate-initialization, so no mem-initializer is involved. None of the exceptions for lifetime extensions apply, therefore the lifetime of that temporary is extended to the lifetime of the reference data member str_.

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auto b = S{"bar"}; // copy-initialization with rvalue

Prior to mandatory copy elision with C++17: Formally, we create a temporary std::string, initialize a temporary S by binding the temporary std::string to the str_ reference member. Then, we move that temporary S into b. This will "copy" the reference, which will not extend the lifetime of the std::string temporary. However, implementations will elide the move from the temporary S to b. This must not affect the lifetime of the temporary std::string though. You can observe this in the following program:

#include <iostream>

#define PRINT_FUNC() { std::cout << __PRETTY_FUNCTION__ << "\n"; }

struct loud
{
    loud() PRINT_FUNC()
    loud(loud const&) PRINT_FUNC()
    loud(loud&&) PRINT_FUNC()
    ~loud() PRINT_FUNC()
};

struct aggr
{
    loud const& l;
    ~aggr() PRINT_FUNC()
};

int main() {
    auto x = aggr{loud{}};
    std::cout << "end of main\n";
    (void)x;
}

Live demo

Note that the destructor of loud is called before the "end of main", whereas x lives until after that trace. Formally, the temporary loud is destroyed at the end of the full-expression which created it.

The behaviour does not change if the move constructor of aggr is user-defined.

With mandatory copy-elision in C++17: We identify the object on the rhs S{"bar"} with the object on the lhs b. This causes the lifetime of the temporary to be extended to the lifetime of b. See CWG 1697.

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For the remaining two examples, the move constructor - if called - simply copies the reference. The move constructor (of S) can be elided, of course, but this is not observable since it only copies the reference.