我有一种情况，我想用一个参数调用一个函数，并将结果返回到同一个参数中

foo = f(foo);

另外，我假设参数x很大，所以我不想调用它的复制构造函数，而是调用它的移动构造函数。最后，我不想通过引用传递参数，因为我想将函数 f 与另一个函数 g 组合在一起。因此，这样的事情

foo = g(f(foo));

有可能。现在，有了移动语义，这一切几乎都是可能的，如下面的程序所示

#include <iostream>

struct Foo {
    Foo() {
        std::cout << "constructor" << std::endl; 
    } 
    Foo(Foo && x) {
        std::cout << "move" << std::endl;
    }
    Foo(Foo const & x) {
        std::cout << "copy" << std::endl;
    }
    ~Foo() {
        std::cout << "destructor" << std::endl;
    }
    Foo & operator = (Foo && x) {
        std::cout << "move assignment" << std::endl; 
        return *this;
    }
    Foo & operator = (Foo & x) {
        std::cout << "copy assignment" << std::endl; 
        return *this;
    }
};

Foo f(Foo && foo) {
    std::cout << "Called f" << std::endl;
    return std::move(foo);
}

Foo g(Foo && foo) {
    std::cout << "Called g" << std::endl;
    return std::move(foo);
}

int main() {
   Foo foo;
   foo = f(std::move(foo));
   std::cout << "Finished with f(foo)" << std::endl;
   foo = g(f(std::move(foo)));
   std::cout << "Finished with g(f(foo))" << std::endl;
}

这个程序的输出是:

constructor
Called f
move
move assignment
destructor
Finished with f(foo)
Called f
move
Called g
move
move assignment
destructor
destructor
Finished with g(f(foo))
destructor

这是有道理的。现在，困扰我的是，当我们第一次调用 f 或组合时，移动构造函数后跟移动赋值运算符。理想情况下，我想使用复制 elison 来防止调用这些构造函数中的任何一个，但我不确定如何调用。具体来说，函数 f 和 g 调用 foo 上的 std::move 因为否则复制，而不是移动，构造函数被调用。这在 C++ 标准的第 12.8.31 和 12.8.32 节中有规定。具体来说，

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the constructor selected for the copy/move operation and/or the destructor for the object have side effects. In such cases, the implementation treats the source and target of the omitted copy/move operation as simply two different ways of referring to the same object, and the destruction of that object occurs at the later of the times when the two objects would have been destroyed without the optimization. This elision of copy/move operations, called copy elision, is permitted in the following circumstances (which may be combined to eliminate multiple copies):

— in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object (other than a function or catch-clause parameter) with the same cvunqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value

因为我们返回一个函数参数，所以我们没有得到复制elison。此外:

When the criteria for elision of a copy operation are met or would be met save for the fact that the source object is a function parameter, and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue. If overload resolution fails, or if the type of the first parameter of the selected constructor is not an rvalue reference to the object’s type (possibly cv-qualified), overload resolution is performed again, considering the object as an lvalue. [ Note: This two-stage overload resolution must be performed regardless of whether copy elision will occur. It determines the constructor to be called if elision is not performed, and the selected constructor must be accessible even if the call is elided. —end note ]

因为我们返回一个函数参数，我们返回一个左值，所以我们被迫使用 std::move。现在，归根结底，我只想将内存移回参数中，同时调用移动构造函数和移动赋值运算符似乎太多了。感觉应该是单招或者copy elison。有办法实现吗？

编辑 1

在对@didierc 的回答的回复中，比评论允许的时间更长，从技术上讲，是的，这适用于这种情况。同时，更大的目标是允许具有多个返回的函数以不复制任何内容的方式组合在一起。我也可以使用移动语义来做到这一点，但它需要 C++14 的技巧才能工作。它也加剧了很多 Action 的问题。但是，从技术上讲，没有拷贝。具体来说:

#include <tuple>
#include <iostream>
#include <utility>

// This comes from the N3802 proposal for C++
template <typename F, typename Tuple, size_t... I>
decltype(auto) apply_impl(F&& f, Tuple&& t, std::index_sequence<I...>) {
    return std::forward<F>(f)(std::get<I>(std::forward<Tuple>(t))...);
}
template <typename F, typename Tuple>
decltype(auto) apply(F&& f, Tuple&& t) {
    using Indices = 
        std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>;
    return apply_impl(std::forward<F>(f), std::forward<Tuple>(t), Indices{});
}

// Now, for our example
struct Foo {
    Foo() {
        std::cout << "constructor" << std::endl; 
    } 
    Foo(Foo && x) {
        std::cout << "move" << std::endl;
    }
    Foo(Foo const & x) {
        std::cout << "copy" << std::endl;
    }
    ~Foo() {
        std::cout << "destructor" << std::endl;
    }
    Foo & operator = (Foo && x) {
        std::cout << "move assignment" << std::endl; 
        return *this;
    }
    Foo & operator = (Foo & x) {
        std::cout << "copy assignment" << std::endl; 
        return *this;
    }
};

std::tuple <Foo,Foo> f(Foo && x,Foo && y) {
    std::cout << "Called f" << std::endl;
    return std::make_tuple <Foo,Foo> (std::move(x),std::move(y));
}

std::tuple <Foo,Foo> g(Foo && x,Foo && y) {
    std::cout << "Called g" << std::endl;
    return std::make_tuple <Foo,Foo> (std::move(x),std::move(y));
}

int main() {
   Foo x,y;
   std::tie(x,y) = f(std::move(x),std::move(y));
   std::cout << "Finished with f(foo)" << std::endl;
   std::tie(x,y) = apply(g,f(std::move(x),std::move(y)));
   std::cout << "Finished with g(f(foo))" << std::endl;
}

这产生

constructor
constructor
Called f
move
move
move assignment
move assignment
destructor
destructor
Finished with f(foo)
Called f
move
move
Called g
move
move
move assignment
move assignment
destructor
destructor
destructor
destructor
Finished with g(f(foo))
destructor
destructor

基本上，出现与上述相同的问题:我们得到移动分配，如果它们消失就好了。

编辑2

根据@MooingDuck 的建议，实际上可以从函数返回一个 rref。通常，这将是一个非常糟糕的主意，但由于内存是在函数外部分配的，因此它不是问题。然后，移动次数急剧减少。不幸的是，如果有人试图将结果分配给 rref，这将导致未定义的行为。所有代码和结果如下。

对于单参数情况:

#include <iostream>

struct Foo {
    // Add some data to see if it gets moved correctly
    int data;

    Foo() : data(0) {
        std::cout << "default constructor" << std::endl; 
    } 
    Foo(int const & data_) : data(data_) {
        std::cout << "constructor" << std::endl; 
    } 
    Foo(Foo && x) {
        data = x.data;
        std::cout << "move" << std::endl;
    }
    Foo(Foo const & x) {
        data = x.data;
        std::cout << "copy" << std::endl;
    }
    ~Foo() {
        std::cout << "destructor" << std::endl;
    }
    Foo & operator = (Foo && x) {
        data = x.data;
        std::cout << "move assignment" << std::endl; 
        return *this;
    }
    Foo & operator = (Foo & x) {
        data = x.data;
        std::cout << "copy assignment" << std::endl; 
        return *this;
    }
};

Foo && f(Foo && foo) {
    std::cout << "Called f: foo.data = " << foo.data << std::endl;
    return std::move(foo);
}

Foo && g(Foo && foo) {
    std::cout << "Called g: foo.data = " << foo.data << std::endl;
    return std::move(foo);
}

int main() {
    Foo foo(5);
    foo = f(std::move(foo));
    std::cout << "Finished with f(foo)" << std::endl;
    foo = g(f(std::move(foo)));
    std::cout << "Finished with g(f(foo))" << std::endl;
    Foo foo2 = g(f(std::move(foo)));
    std::cout << "Finished with g(f(foo)) a second time" << std::endl;
    std::cout << "foo2.data = " << foo2.data << std::endl;
    // Now, break it.
    Foo && foo3 = g(f(Foo(4)));  
    // Notice that the destuctor for Foo(4) occurs before the following line.
    // That means that foo3 points at destructed memory.
    std::cout << "foo3.data = " << foo3.data << ".  If there's a destructor"
        " before this line that'd mean that this reference is invalid."
        << std::endl;
}

这产生

constructor
Called f: foo.data = 5
move assignment
Finished with f(foo)
Called f: foo.data = 5
Called g: foo.data = 5
move assignment
Finished with g(f(foo))
Called f: foo.data = 5
Called g: foo.data = 5
move
Finished with g(f(foo)) a second time
foo2.data = 5
constructor
Called f: foo.data = 4
Called g: foo.data = 4
destructor
foo3.data = 4.  If there's a destructor before this line that'd mean that this reference is invalid.
destructor
destructor

在多参数情况下

#include <tuple>
#include <iostream>
#include <utility>

// This comes from the N3802 proposal for C++
template <typename F, typename Tuple, size_t... I>
decltype(auto) apply_impl(F&& f, Tuple&& t, std::index_sequence<I...>) {
    return std::forward<F>(f)(std::get<I>(std::forward<Tuple>(t))...);
}
template <typename F, typename Tuple>
decltype(auto) apply(F&& f, Tuple&& t) {
    using Indices = 
        std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>;
    return apply_impl(std::forward<F>(f), std::forward<Tuple>(t), Indices{});
}

// Now, for our example
struct Foo {
    // Add some data to see if it gets moved correctly
    int data;

    Foo() : data(0) {
        std::cout << "default constructor" << std::endl; 
    } 
    Foo(int const & data_) : data(data_) {
        std::cout << "constructor" << std::endl; 
    } 
    Foo(Foo && x) {
        data = x.data;
        std::cout << "move" << std::endl;
    }
    Foo(Foo const & x) {
        data = x.data;
        std::cout << "copy" << std::endl;
    }
    ~Foo() {
        std::cout << "destructor" << std::endl;
    }
    Foo & operator = (Foo && x) {
        std::cout << "move assignment" << std::endl; 
        return *this;
    }
    Foo & operator = (Foo & x) {
        std::cout << "copy assignment" << std::endl; 
        return *this;
    }
};

std::tuple <Foo&&,Foo&&> f(Foo && x,Foo && y) {
    std::cout << "Called f: (x.data,y.data) = (" << x.data << ',' <<
        y.data << ')' << std::endl;
    return std::tuple <Foo&&,Foo&&> (std::move(x),std::move(y));
}

std::tuple <Foo&&,Foo&&> g(Foo && x,Foo && y) {
    std::cout << "Called g: (x.data,y.data) = (" << x.data << ',' <<
        y.data << ')' << std::endl;
    return std::tuple <Foo&&,Foo&&> (std::move(x),std::move(y));
}

int main() {
    Foo x(5),y(6);
    std::tie(x,y) = f(std::move(x),std::move(y));
    std::cout << "Finished with f(x,y)" << std::endl;
    std::tie(x,y) = apply(g,f(std::move(x),std::move(y)));
    std::cout << "Finished with g(f(x,y))" << std::endl;
    std::tuple <Foo,Foo> x_y = apply(g,f(std::move(x),std::move(y)));
    std::cout << "Finished with g(f(x,y)) a second time" << std::endl;
    std::cout << "(x.data,y.data) = (" << std::get <0>(x_y).data << ',' <<
        std::get <1> (x_y).data << ')' << std::endl;
    // Now, break it.
    std::tuple <Foo&&,Foo&&> x_y2 = apply(g,f(Foo(7),Foo(8)));  
    // Notice that the destuctors for Foo(7) and Foo(8) occur before the
    // following line.  That means that x_y2points at destructed memory.
    std::cout << "(x2.data,y2.data) = (" << std::get <0>(x_y2).data << ',' <<
        std::get <1> (x_y2).data << ')' << ".  If there's a destructor"
        " before this line that'd mean that this reference is invalid."
        << std::endl;
}

这产生

constructor
constructor
Called f: (x.data,y.data) = (5,6)
move assignment
move assignment
Finished with f(x,y)
Called f: (x.data,y.data) = (5,6)
Called g: (x.data,y.data) = (5,6)
move assignment
move assignment
Finished with g(f(x,y))
Called f: (x.data,y.data) = (5,6)
Called g: (x.data,y.data) = (5,6)
move
move
Finished with g(f(x,y)) a second time
(x.data,y.data) = (5,6)
constructor
constructor
Called f: (x.data,y.data) = (7,8)
Called g: (x.data,y.data) = (7,8)
destructor
destructor
(x2.data,y2.data) = (7,8).  If there's a destructor before this line that'd mean that this reference is invalid.
destructor
destructor
destructor
destructor

最佳答案

在我看来，您需要的不是一种通过函数调用来回移动值的机制，因为引用足以做到这一点，而是一种以这种方式组合函数的设备。

template <void f(Foo &), void g(Foo &)>
void compose2(Foo &v){
   f(v);
   g(v);
}

当然，您可以在参数类型上使其更通用。

template <typename T, void f(T&), void (...G)(T&)>
void compose(T &v){
  f(v);
  compose2<T,G...>(v);
}

template <typename T>
void compose(Foo &){
}

例子:

#include <iostream>

//... above template definitions for compose elided


struct Foo {
  int x;
};

void f(Foo &v){
  v.x++;
}

void g(Foo &v){
  v.x *= 2;
}

int main(){
  Foo v = { 9 };

  compose<Foo, f, g, f, g>(v);

  std::cout << v.x << "\n"; // output "42"
}

请注意，您甚至可以在过程原型(prototype)上参数化模板，但此时在我的机器上，似乎只有 clang++ (v3.5) 接受它，g++ (4.9.1) 不喜欢它。

关于c++ - 当复制 elison 失败时，有没有办法阻止移动构造函数后跟移动赋值运算符？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28412305/