copy-/move-constructors的奇怪行为以及如何返回大对象？

最近，在缺席一段时间后，我再次尝试使用C++11。在阅读了互联网上的许多文章后，我现在完全不知道从工厂函数（基本上是从数据库中进行数据分析）返回大型对象的最有效方法是什么。

我已经成为unique_ptr的粉丝，但我在几篇文章中读到，由于有了新的move构造函数，现在完全可以通过值返回一个大向量，比如，并且由于这些新的语义，应该像复制一个指针一样快。

为了尝试这一点，我编写了一个小测试程序，在各种构造函数中都有输出：

#include <iostream>
#include <memory>
using namespace std;

class C {
public:
    C( string n ) : _name{n} { cout << "Constructing a C named '" << _name << "'n"; };
    C() : _name( "EMPTY" ) { cout << "Default-constructing a C named '" << _name << "'n"; } ;     // default-ctor
    C( const C& c ) : _name{c._name} {
        _name += " [copied]";
        cout << "Copy-constructing a C named '" << _name << "'n";
    };
    C( C&& c )
        : _name{c._name} {
        _name += " [moved]";
        cout << "Move-constructing a C named '" << _name << "'n";
    };
    ~C() { cout << "Destructing a C named '" << _name << "'n"; };
    string getName() { return _name; };
private:
    string _name;
};

并用测试

C fooVal() {    
    cout << "In fooValn";
    string str = "value return";
    C c(str);
    return c;
}
C& fooRef() {
    cout << "In fooRefn";
    string str = "reference return";
    C* pC = new C( str );
    return *pC;
}
C* fooPtr() {
    cout << "In fooPtrn";
    string str = "classical pointer return";
    C* pC = new C( str );
    return pC;
}
unique_ptr<C> fooUPtr() {
    cout << "In fooUPtrn";
    string str = "unique_ptr return";
    return unique_ptr<C>(new C(str));
}
shared_ptr<C> fooSPtr() {
    cout << "In fooSPtrn";
    string str = "shared_ptr return";
    return shared_ptr<C>(new C(str));
}
// IMPORTANT: THIS NEEDS TO BE COMPILED WITH FLAG -fno-elide-constructors
int main(int argc, const char * argv[])
{
    C cv(fooVal());        
    cout << "cv constructedn";    
    C& cr = fooRef();        
    cout << "cr constructedn";        
    C* pC = fooPtr();        
    cout << "*pC constructedn";        
    unique_ptr<C> upC = fooUPtr();        
    cout << "*upC constructedn";        
    shared_ptr<C> spC = fooSPtr();        
    cout << "*spC constructedn";        
    cout << "Alive: " << cv.getName() << ", " << cr.getName() << ", " << pC->getName() << ", " << upC->getName() << ".n";
}

现在，如果我按原样编译它，编译器会优化（"elides"）各种构造函数调用，我会得到输出：

In fooVal
Constructing a C named 'value return'
cv constructed
In fooRef
Constructing a C named 'reference return'
cr constructed
In fooPtr
Constructing a C named 'classical pointer return'
*pC constructed
In fooUPtr
Constructing a C named 'unique_ptr return'
*upC constructed
In fooSPtr
Constructing a C named 'shared_ptr return'
*spC constructed
Alive: value return, reference return, classical pointer return, unique_ptr return.
Destructing a C named 'shared_ptr return'
Destructing a C named 'unique_ptr return'
Destructing a C named 'value return'

好吧，但是你可以看到有多少复制被优化掉了。为了了解"指定"行为是什么，我使用标志-fno-elide构造函数编译了它（我使用的是Apple LLVM 4.2版（clang-425.0.28））

In fooVal
Constructing a C named 'value return'
Destructing a C named 'value return'
Move-constructing a C named ' [moved]'
Destructing a C named ''
cv constructed
In fooRef
Constructing a C named 'reference return'
cr constructed
In fooPtr
Constructing a C named 'classical pointer return'
*pC constructed
In fooUPtr
Constructing a C named 'unique_ptr return'
*upC constructed
In fooSPtr
Constructing a C named 'shared_ptr return'
*spC constructed
Alive:  [moved], reference return, classical pointer return, unique_ptr return.
Destructing a C named 'shared_ptr return'
Destructing a C named 'unique_ptr return'
Destructing a C named ' [moved]'

所以，很明显，返回值的对象发生了一些可疑的事情。显然，这不仅仅是一个小问题，因为我本以为-fno-elide构造函数不会改变语义，只会改变所涉及的构造函数数量。

因此我问：

1. 发生了什么事？为什么value对象会"丢失"其字符串参数？在哪里
2. 看起来价值回报有问题，而其他回报运作良好。那么，为什么现在人们建议我们按价值回归，"系统会照顾剩下的"呢
3. 返回大型对象的好方法是什么
4. 我是不是在一个我没有看到的地方犯了一个错误

谢谢！