两步复制省略以捕获构造函数调用中的右值作为实例变量

Question

I am trying to get an rvalue instance of this class:

#include <iostream>
#define msg(x) std::cout << x " constructor\n"

struct X {
    int i;
    X(int i) : i(i) {msg("X");}
    X(const X& x) : i(x.i) {std::cout << "X copy\n";}
    X(X&& x) {std::swap(i, x.i); std::cout << "X move\n";}
};

into instance variable x of this class:

struct A {
    X x;
    A(X x) : x(x) {msg("A");}
};

like so:

int main() {
    A a(X(1));
    std::cout << a.x.i << "\n\n";
}

without any copies or moves being made .

According to these references,

• http://thbecker.net/articles/rvalue_references/section_01.html
• http://www.slideshare.net/oliora/hot-11-2-new-style-arguments-passing

and many many posts on SO ( so please read to the end before flagging as duplicate ), I should rely on copy elision , whose conditions should be satisfied if I pass by value . Note that there are two copy elisions required, namely:

constructor call -> constructor local variable -> instance variable

as can be seen when turning copy elision off (compile with g++-4.8 -std=c++11 -fno-elide-constructors ):

X constructor
X move
X copy
A constructor
1

So there is one move step and one copy step, which should both go away if I turn copy elision on (compile with g++-4.8 -std=c++11 -O3 ):

X constructor
X copy
A constructor
1

Bummer, the copy step remained!

---

Can I get any better with any other variation of std::move() , std::forward or passing as rvalue-reference ?

struct B {
    X x;
    B(X x) : x(std::move(x)) {msg("B");}
};

struct C {
    X x;
    C(X x) : x(std::forward<X>(x)) {msg("C");}
};

struct D {
    X x;
    D(X&& x) : x(std::move(x)) {msg("D");}
};

int main() {
    B b(X(2));
    std::cout << b.x.i << "\n\n";

    C c(X(3));
    std::cout << c.x.i << "\n\n";

    D d(X(4));
    std::cout << d.x.i << "\n\n";
}

which produces the output:

X constructor
X move
B constructor
2

X constructor
X move
C constructor
3

X constructor
X move
D constructor
4

OK, I turned the copy into a move , but this is not satisfactory!

---

Next, I tried to make the instance variable x a reference X& :

struct E {
    X& x;
    E(X x) : x(x) {msg("E");}
};

int main() {
    E e(X(5));
    std::cout << e.x.i << "\n\n";
}

which produces:

X constructor
E constructor
1690870696

Bad idea! I got rid of the move but the rvalue instance that x was referencing to got destroyed under my seat, so the last line prints garbage instead of 5 . Two notes:

• g++-4.8 didn't warn me of anything, even with -pedantic -Wall -Wextra
• The program prints 5 when compiled with -O0

So this bug may go unnoticed for quite a while!

---

So, is this a hopeless case? Well no:

struct F {
    X& x;
    F(X& x) : x(x) {msg("F");}
};

int main() {
    X x(6);
    F f(x);
    std::cout << f.x.i << "\n";
}

prints:

X constructor
F constructor
6

Really? No fancy new C++11 features, no copy elision at the discretion of the compiler, just plain old FORTRAN66-style pass-by-reference does what I want and probably will perform best?

So here are my questions:

• Is there any way one can get this to work for rvalues ? Did I miss any features?
• Is the lvalue -reference version really the best, or are there hidden costs in the X x(6) step?
• Could there be any inconveniences introduced by x living on after the construction of f ?
• Could I pay a data locality penalty for using the lvalue reference to an external instance?

Answer 1

Without going into too much detail of your question, copy elision is basically used as much as possible. Here's a quick demo:

#include <iostream>
#include <utility>

struct X
{
    int n_;

    explicit X(int n) : n_(n) { std::cout << "Construct: " << n_ << "\n"; }
    X(X const & rhs) : n_(rhs.n_) { std::cout << "X copy:" << n_ << "\n"; }
    X(X && rhs) : n_(rhs.n_) { rhs.n_ = -1; std::cout << "X move:" << n_ << "\n"; }
   ~X() { std::cout << "Destroy: " << n_ << "\n"; }
};

struct A
{
    explicit A(X x) : x_(std::move(x)) {};
    X x_;
};

struct B
{
    X x;
};

int main()
{
    A a(X(12));
    B b { X(24) };
}

This produces :

Construct: 12
X move:12
Destroy: -1
Construct: 24
Destroy: 24
Destroy: 12

The one move in x_(std::move(x)) is not elidable, since it doesn't involve a function return. But that's pretty good anyway. And notice how the aggregate b is indeed initialized "in-place".

---

Your example F shows that you're willing to expose the coupling of X to its ambient class. In that case, you could make a special constructor for A that constructs the X directly:

struct A
{
    explicit A(X x) : x_(std::move(x)) {};
    X x_;

    // Better approach
    struct direct_x_t {};
    static direct_x_t direct_x;

    // In our case, this suffices:
    A(direct_x_t, int n) : x_(n) {}

    // Generally, a template may be better: (TODO: add SFINAE control)
    template <typename ...Args> A(direct_x_t, Args &&... args)
    : x_(std::forward<Args>(args)...) {}
};

Usage:

A a(A::direct_x, 36);