当包含lambda的类型被赋值时会发生什么？ [英] What happens when a type containing a lambda gets assigned?

问题描述

说我有某种类型的包装函数，可能是lambda函数：

Say I have some sort of type wrapping up a function, maybe a lambda function:

template<typename Function>
  struct my_struct
{
  Function f;

  my_struct(const Function &f) : f(f) {}
};

当分配此类型的实例时会发生什么？我的理解是，lambdas是不可变的，并且已经删除了赋值运算符。

What happens when an instance of this type is assigned? My understanding is that lambdas are immutable, and have deleted assignment operators.

然而，当我在下面的代码片段中指定一个对象时，不会发出错误：

Yet, when I assign to an object this type in the code snippet below, no error is emitted:

// a structure which contains a function;
// possibly a lambda function
template<typename Function>
  struct my_struct
{
  Function f;

  my_struct(const Function &f) : f(f) {}

  // XXX adding this assignment operator causes an error
  //my_struct &operator=(const my_struct &other)
  //{
  //  f = other.f;
  //  return *this;
  //}
};

template<typename Function>
my_struct<Function> make_struct(const Function &f)
{
  return my_struct<Function>(f);
}

int main()
{
  // create some lambda
  auto lambda = [](int x){return x;};

  // make a struct containing a copy of the lambda
  auto x = make_struct(lambda);

  // try to assign to the struct, which
  // presumably assigns to the enclosed lambda
  x = make_struct(lambda);

  return 0;
}

添加注释的赋值运算符会产生错误， p>

Adding the commented-out assignment operator yields an error, as expected:

$ g++-4.6 -std=c++0x test.cpp
test.cpp: In member function ‘my_struct<Function>& my_struct<Function>::operator=(const my_struct<Function>&) [with Function = main()::<lambda(int)>, my_struct<Function> = my_struct<main()::<lambda(int)> >]’:
test.cpp:34:25:   instantiated from here
test.cpp:13:5: error: use of deleted function ‘main()::<lambda(int)>& main()::<lambda(int)>::operator=(const main()::<lambda(int)>&)’
test.cpp:27:18: error: a lambda closure type has a deleted copy assignment operator

成员变量？这似乎是一个合理的事情想试试。例如，考虑组合一个lambda和 boost :: transform_iterator 。

So, is it possible to create assignable types with lambda member variables? This seems like a reasonable thing to want to try. Consider combining a lambda with boost::transform_iterator, for example.

推荐答案

你就近了。 lambda具有隐式复制构造函数，并且可能具有 - 取决于捕获的值 - 隐式移动构造函数。它有一个删除的副本赋值运算符。

You're close. A lambda has an implicit copy-constructor and may have — depending on the captured values — an implicit move-constructor. It has a deleted copy-assignment operator.

换句话说，你可以构造它，但你不能分配它。如果你正在寻找一个泛型函数对象，你想使用 std :: function<> 。

In other words, you may construct it, but you may not assign it. If you're looking for a generic function object, you want to use std::function<>. It emulates functions as first-class values.

请注意，immutable不同于assignable。当一个lambda被称为mutable时，这意味着它的函数调用体可以修改lambda的成员（即函数不是 const ）：

Note that immutable is different from assignable. When a lambda is called mutable, that means its function-call body can modify the members of the lambda (i.e., the function is not const):

int i = 0;

auto okay = [=](int x) mutable { i += x; };
auto error = [=](int x) { i += x; };

这些都是可复制构造且不可分配。

Each of these is copy-constructable and non-assignable.

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