如何捕获列表的lambda实际工作在C ++ 11？ [英] How do capture lists of lambdas actually work in C++11?

问题描述

我知道捕获列表使变量在lambda函数体内可用，如下所示：

I know that capture lists make variables available inside a lambda function body like so:

int pos(0);
std::function<void()> incPos = [&pos](){ ++pos; };
incPos(); //pos is now 1

但是这种捕获如何在编译器级别上工作呢？

But how does that capturing actually work on compiler level? Where are the captured addresses or captured values stored?

推荐答案

每个lambda表达式都会生成一个唯一的函数对象变量作为数据成员。例如，你的代码中的lambda表达式将被编译器转换成这样：

Each lambda expression generates a unique function object (closure) that stores the captured variables as data members. For instance, the lambda expression in your code would be transformed into something like this by the compiler:

struct __uniquely_named_lambda
{
  __uniquely_named_lambda(int& pos)
  : pos(pos) {}
  int& pos;

  void operator()() const
  { ++pos; }
};

调用lambda只是调用 operator $ c>。

Invoking the lambda is simply a call to operator().

数据成员是引用，因为您通过引用捕获。如果你通过值捕获它将是一个简单的 int 。另请注意，默认情况下生成的 operator（）为 const 。这就是为什么你不能修改捕获的变量，除非你使用 mutable 关键字。

The data member is a reference since you captured by reference. If you captured by value it would be a plain int. Also note that generated operator() is const by default. This is why you cannot modify captured variables unless you use the mutable keyword.

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