如何在编译时检测类型是否为lambda表达式? [英] How to detect whether a type is a lambda expression at compile time?

问题描述

假设我有一个my_struct类型，其中包含一个成员变量f，它是一个函数. f可能是c ++ 11 lambda函数.

Suppose I have a type my_struct enclosing a member variable, f, which is a function. It's possible for f to be a c++11 lambda function.

由于分配给lambda对象是非法的，所以我想以my_struct的赋值运算符实现这种方式，即当f是lambda时，不对其进行分配.

Since it is illegal to assign to lambda objects, I'd like to implement my_struct's assignment operator in such a way that when f is a lambda, it is not assigned.

是否可以构建类型特征is_lambda来检查类型的lambda值?

Is it possible to build a type trait is_lambda which can inspect a type for lambda-ness?

在代码中:

#include <type_traits>

template<typename Function> struct is_lambda
{
  // what goes here?
};

template<typename Function> struct my_struct
{
  Function f;

  my_struct &do_assign(const my_struct &other, std::true_type)
  {
    // don't assign to f
    return *this;
  }

  my_struct &do_assign(const my_struct &other, std::false_type)
  {
    // do assign to f
    f = other.f;
    return *this;
  }

  my_struct &operator=(const my_struct &other)
  {
    return do_assign(other, typename is_lambda<Function>::type());
  }
};

推荐答案

在没有编译器支持的情况下是不可能的，因为lambda的类型只是普通的非工会类类型.

Impossible without compiler support, as the type of a lambda is just a normal, non-union class type.

§5.1.2 [expr.prim.lambda] p3

lambda-expression 的类型(也是闭包对象的类型)是唯一的，未命名的ununion类类型[...]

The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion class type [...]

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