Lambda 函数作为基类 [英] Lambda functions as base classes

问题描述

在使用 Lambda 时，我发现了一个我不完全理解的有趣行为.

Playing around with Lambdas I found an interesting behaviour that I do not fully understand.

假设我有一个从 2 个模板参数派生的 struct Overload，并且有一个 using F1::operator(); 子句.

Supose I have a struct Overload that derives from 2 template parameters, and has a using F1::operator(); clause.

现在如果我从两个函子派生，我只能访问 F1 的 operator()(正如我所期望的)

Now if I derive from two functors I can only access the operator() of F1 (as I would expect)

如果我从两个 Lambda 函数派生，这将不再正确:我也可以从 F2 访问 operator().

If I derive from two Lambda Functions this is no longer true: I can access the operator() from F2 too.

#include <iostream>

// I compiled with g++ (GCC) 4.7.2 20121109 (Red Hat 4.7.2-8)
//
// g++ -Wall -std=c++11 -g main.cc
// g++ -Wall -std=c++11 -DFUNCTOR -g main.cc
// 
// or clang clang version 3.3 (tags/RELEASE_33/rc2)
// 
// clang++ -Wall -std=c++11 -g main.cc
// clang++ -Wall -std=c++11 -DFUNCTOR -g main.cc
// 
// on a Linux localhost.localdomain 3.9.6-200.fc18.i686 #1 SMP Thu Jun 13 
// 19:29:40 UTC 2013 i686 i686 i386 GNU/Linux box


struct Functor1
{
    void operator()() { std::cout << "Functor1::operator()()
"; }
};

struct Functor2
{
    void operator()(int) { std::cout << "Functor2::operator()(int)
"; }
};

template <typename F1, typename F2>
struct Overload : public F1, public F2
{
    Overload()
        : F1()
        , F2() {}

    Overload(F1 x1, F2 x2)
        : F1(x1)
        , F2(x2) {}

    using F1::operator(); 
};

template <typename F1, typename F2>
auto get(F1 x1, F2 x2) -> Overload<F1, F2>
{
   return Overload<F1, F2>(x1, x2);
}


int main(int argc, char *argv[])
{
    auto f = get(Functor1(), Functor2());

    f();
#ifdef FUNCTOR
    f(2); // this one doesn't work IMHO correctly
#endif

    auto f1 = get(
                  []() { std::cout << "lambda1::operator()()
"; },
                  [](int) { std::cout << "lambda2::operator()(int)
"; }
                  );
    f1();
    f1(2); // this one works but I don't know why


  return 0;
}

标准规定:

lambda 表达式的类型(也是闭包对象的类型)是唯一的、未命名的非联合类类型

The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non- union class type

所以每个 Lambda 的类型都应该是唯一的.

So every Lambda's types should be unique.

我无法解释为什么会这样:有人能解释一下吗?

I cannot explain why this is so: can anyone shed some light on this please?

推荐答案

除了 operator() 之外，由 lambda 定义的类可以(在适当的情况下)提供到指向函数的指针.情况(或至少是主要情况)是 lambda 无法捕获任何内容.

In addition to operator(), a the class defined by a lambda can (under the right circumstances) provide a conversion to a pointer to function. The circumstance (or at least the primary one) is that the lambda can't capture anything.

如果您添加捕获:

auto f1 = get(
              []() { std::cout << "lambda1::operator()()
"; },
              [i](int) { std::cout << "lambda2::operator()(int)
"; }
              );
f1();
f1(2);

...不再提供到 pointer to function 的转换，因此尝试编译上面的代码会给出您可能一直期望的错误:

...the conversion to pointer to function is no longer provided, so trying to compile the code above gives the error you probably expected all along:

trash9.cpp: In function 'int main(int, char**)':
trash9.cpp:49:9: error: no match for call to '(Overload<main(int, char**)::<lambda()>, main(int, char**)::<lambda(int)> >) (int)'
trash9.cpp:14:8: note: candidate is:
trash9.cpp:45:23: note: main(int, char**)::<lambda()>
trash9.cpp:45:23: note:   candidate expects 0 arguments, 1 provided

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