问题描述
我正在努力赶上 C++11 和所有很棒的新功能.我有点坚持 lambdas.
I'm trying to catch up on C++11 and all the great new features. I'm a bit stuck on lambdas.
这是我能够开始工作的代码:
Here's the code I was able to get to work:
#include <iostream>
#include <cstdlib>
#include <vector>
#include <string>
#include <functional>
using namespace std;
template<typename BaseT, typename Func>
vector<BaseT> findMatches(vector<BaseT> search, Func func)
{
vector<BaseT> tmp;
for(auto item : search)
{
if( func(item) )
{
tmp.push_back(item);
}
}
return tmp;
}
void Lambdas()
{
vector<int> testv = { 1, 2, 3, 4, 5, 6, 7 };
auto result = findMatches(testv, [] (const int &x) { return x % 2 == 0; });
for(auto i : result)
{
cout << i << endl;
}
}
int main(int argc, char* argv[])
{
Lambdas();
return EXIT_SUCCESS;
}
我想要的是这个:
template<typename BaseT>
vector<BaseT> findMatches(vector<BaseT> search, function <bool (const BaseT &)> func)
{
vector<BaseT> tmp;
for(auto item : search)
{
if( func(item) )
{
tmp.push_back(item);
}
}
return tmp;
}
基本上我想将可能的 lambdas 缩小到一个合理的函数子集.我错过了什么?这甚至可能吗?我使用的是 GCC/G++ 4.6.
Basically I want to narrow down the possible lambdas to a sensible subset of functions. What am I missing? Is this even possible? I'm using GCC/G++ 4.6.
推荐答案
Stephan T. Lavavej 解释了为什么这在 这个视频.基本上，问题在于编译器试图从 both std::vector 和 推导出 BaseTcode>std::function 参数.C++ 中的 lambda 不是 std::function 类型，它是一种未命名的、唯一的非联合类型，如果它没有捕获列表(空 []).另一方面，可以从任何可能类型的可调用实体(函数指针、成员函数指针、函数对象)创建 std::function 对象.
Stephan T. Lavavej explains why this doesn't work in this video. Basically, the problem is that the compiler tries to deduce BaseT from both the std::vector and the std::function parameter. A lambda in C++ is not of type std::function, it's an unnamed, unique non-union type that is convertible to a function pointer if it doesn't have a capture list (empty []). On the other hand, a std::function object can be created from any possible type of callable entity (function pointers, member function pointers, function objects).
请注意，我个人不明白为什么您要将传入的函子限制为该特定签名(除了通过多态函数包装器间接进行的事实之外，例如 std::function, 远比直接调用函子(甚至可能被内联)低效)，但这是一个工作版本.基本上，它禁用了 std::function 部分的参数推导，并且只从 std::vector 参数推导了 BaseT:
Note that I personally don't understand why you would want to limit the incoming functors to that specific signature (in addition to the fact that indirection through a polymorphic function wrapper, like std::function, is by far more inefficient than a direct call to a functor (which may even be inlined)), but here's a working version. Basically, it disables argument deduction on the std::function part, and only deduces BaseT from the std::vector argument:
template<class T>
struct Identity{
typedef T type;
};
template<typename BaseT>
vector<BaseT> findMatches(vector<BaseT> search,
typename Identity<function<bool (const BaseT &)>>::type func)
{
vector<BaseT> tmp;
for(auto item : search)
{
if( func(item) )
{
tmp.push_back(item);
}
}
return tmp;
}
Ideone 上的实例.
另一种可能的方法是不直接限制函子类型，而是通过 SFINAE 间接限制:
Another possible way would be to not restrict the functor type directly, but indirectly through SFINAE:
template<class T, class F>
auto f(std::vector<T> v, F fun)
-> decltype(bool(fun(v[0])), void())
{
// ...
}
Ideone 上的实例.
如果 fun 不接受 T& 类型的参数或者返回类型不能转换为 ，这个函数将从重载集中删除布尔., void() 使得 f 的返回类型为 void.
This function will be removed from the overload set if fun doesn't take an argument of type T& or if the return type is not convertible to bool. The , void() makes f's return type void.
这篇关于Lambda 和 std::function的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！