C++ 错误：使用删除的功能。为什么？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26749505/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
error: use of deleted function. Why?
提问by quant
I am trying to create a function that applies an arbitrary functor Fto every element of a provided tuple:
我正在尝试创建一个将任意函子应用于F提供的元组的每个元素的函数：
#include <functional>
#include <tuple>
// apply a functor to every element of a tuple
namespace Detail {
template <std::size_t i, typename Tuple, typename F>
typename std::enable_if<i != std::tuple_size<Tuple>::value>::type
ForEachTupleImpl(Tuple& t, F& f)
{
f(std::get<i>(t));
ForEachTupleImpl<i+1>(t, f);
}
template <std::size_t i, typename Tuple, typename F>
typename std::enable_if<i == std::tuple_size<Tuple>::value>::type
ForEachTupleImpl(Tuple& t, F& f)
{
}
}
template <typename Tuple, typename F>
void ForEachTuple(Tuple& t, F& f)
{
Detail::ForEachTupleImpl<0>(t, f);
}
struct A
{
A() : a(0) {}
A(A& a) = delete;
A(const A& a) = delete;
int a;
};
int main()
{
// create a tuple of types and initialise them with zeros
using T = std::tuple<A, A, A>;
T t;
// creator a simple function object that increments the objects member
struct F
{
void operator()(A& a) const { a.a++; }
} f;
// if this works I should end up with a tuple of A's with members equal to 1
ForEachTuple(t, f);
return 0;
}
Live code example: http://ideone.com/b8nLCy
实时代码示例：http: //ideone.com/b8nLCy
I don't want to create copies of Abecause it might be expensive (obviously in this example it is not) so I deleted the copy constructor. When I run the above program I get:
我不想创建 的副本，A因为它可能很昂贵（在这个例子中显然不是）所以我删除了复制构造函数。当我运行上述程序时，我得到：
/usr/include/c++/4.8/tuple:134:25: error: use of deleted function ‘A::A(const A&)' : _M_head_impl(__h) { }
/usr/include/c++/4.8/tuple:134:25: error: use of deleted function ‘A::A(const A&)' : _M_head_impl(__h) { }
I know that the constructor is deleted (that was intentional) but what I don't understand is why it is trying to make copies of my struct. Why is this happening, and how can I achieve this without copying A?
我知道构造函数被删除了（这是故意的），但我不明白的是为什么它试图复制我的结构。为什么会发生这种情况，我如何在不复制的情况下实现这一目标A？
回答by Ben Voigt
This is the problem you're getting a "deleted constructor" error for:
这是您收到“已删除构造函数”错误的问题：
std::function<void(A)> f = [](A& a) { a.a++; };
You're trying to set up a std::functionthat passes an Aby value. But A, having no copy-constructor, can't be passed by value.
您正在尝试设置std::function一个A通过值传递的。但是A，没有复制构造函数，不能按值传递。
Try matching the actual argument type more carefully:
尝试更仔细地匹配实际参数类型：
std::function<void(A&)> f = [](A& a) { a.a++; };
But since you aren't capturing variables, you can simply try
但是由于您没有捕获变量，您可以简单地尝试
void(*f)(A&) = [](A& a) { a.a++; };
You've also got a major problem with the base case of your template recursion: even if you get enable_ifworking, which it seems not to be, you'll have an ambiguous call. I think you need to also disablethe main case.
您的模板递归的基本情况也有一个主要问题：即使您开始enable_if工作（似乎并非如此），您也会遇到模棱两可的调用。我认为您还需要禁用主要案例。
回答by Dietmar Kühl
When instantiating std::function<void(A)>a function with a signature taking an Aby value is instantiated. To synthesize this function definition a copy or a move of Ais needed. Since Ais non-copyable but not made movable, such a constructor doesn't exist.
当实例化std::function<void(A)>一个带有A按值的签名的函数时被实例化。要综合此函数定义，需要复制或移动A。由于A不可复制但不可移动，因此不存在这样的构造函数。
Of course, there is also this funny thing:
当然，也有这么搞笑的事情：
T t();
which declares a function return a Tby value requiring a copy. When calling ForEachTuple(t, f)this function is referenced and instantiated. You may want to replace this declaration by one of
它声明一个函数返回一个T需要副本的值。调用ForEachTuple(t, f)此函数时会被引用和实例化。您可能希望用以下之一替换此声明
T t;
T t{};
(in the answer above I had only looked at the first problem I spotted).
（在上面的答案中，我只看了我发现的第一个问题）。
