Lambda表达式作为类模板参数 [英] Lambda expressions as class template parameters

问题描述

可以将lambda表达式用作类模板参数吗？ （请注意，这是一个与这一个非常不同的问题，它询问是否lambda表达式本身可以模板化。）

Can lambda expressions be used as class template parameters? (Note this is a very different question than this one, which asks if a lambda expression itself can be templated.)

我问你是否可以这样做：

I'm asking if you can do something like:

template <class Functor> 
struct Foo { };
// ...
Foo<decltype([]()->void { })> foo;

这在例如类模板有各种参数例如 equal_to 或者什么，通常实现为一线函数。例如，假设我想实例化一个使用我自己的自定义等式比较函数的哈希表。我想能够说一些像：

This would be useful in cases where, for example, a class template has various parameters like equal_to or something, which are usually implemented as one-liner functors. For example, suppose I want to instantiate a hash table which uses my own custom equality comparison function. I'd like to be able to say something like:

typedef std::unordered_map<
  std::string,
  std::string,
  std::hash<std::string>,
  decltype([](const std::string& s1, const std::string& s2)->bool 
    { /* Custom implementation of equal_to */ })
  > map_type;

但是我在GCC 4.4和4.6上测试了它，它不工作，显然是因为匿名由lambda表达式创建的类型没有默认构造函数。 （我记得一个类似的问题与 boost :: bind 。）有一些原因草案标准不允许这个，或者我错了，它是允许的，但GCC

But I tested this on GCC 4.4 and 4.6, and it doesn't work, apparently because the anonymous type created by a lambda expression doesn't have a default constructor. (I recall a similar issue with boost::bind.) Is there some reason the draft standard doesn't allow this, or am I wrong and it is allowed but GCC is just behind in their implementation?

推荐答案

我问你是否可以做一些类似的工作： / p>

I'm asking if you can do something like:

Foo<decltype([]()->void { })> foo;

不能，因为lambda表达式不会出现在未评估的上下文中c $ c> decltype 和 sizeof 等）。
C ++ 0x FDIS，5.1.2 [expr.prim.lambda] p2

No you can't, because lambda expressions shall not appear in an unevaluated context (such as decltype and sizeof, amongst others). C++0x FDIS, 5.1.2 [expr.prim.lambda] p2

导致临时的临时值（12.2）。这个临时变量称为
闭包对象。 lambda表达式不应出现在未求值的操作数中（第5条）。 [注意：A
闭包对象的行为类似于一个函数对象（20.8）.- end note]
（强调我）

The evaluation of a lambda-expression results in a prvalue temporary (12.2). This temporary is called the closure object. A lambda-expression shall not appear in an unevaluated operand (Clause 5). [ Note: A closure object behaves like a function object (20.8).—end note ] (emphasis mine)

您需要首先创建一个特定的lambda，然后使用decltype：

You would need to first create a specific lambda and then use decltype on that:

auto my_comp = [](const std::string& left, const std::string& right) -> bool {
  // whatever
}

typedef std::unordered_map<
  std::string,
  std::string,
  std::hash<std::string>,
  decltype(my_comp)
  > map_type;

这是因为每个lambda派生的闭包对象可以有一个完全不同的类型， em> anonymous 功能。

That is because each lambda-derived closure object could have a completely different type, they're like anonymous functions after all.

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