Lambda 函数作为基类
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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