問題描述
我正在努力趕上 C++11 和所有很棒的新功能.我有點(diǎn)堅(jiān)持 lambdas.
I'm trying to catch up on C++11 and all the great new features. I'm a bit stuck on lambdas.
這是我能夠開始工作的代碼:
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;
}
我想要的是這個(gè):
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;
}
基本上我想將可能的 lambdas 縮小到一個(gè)合理的函數(shù)子集.我錯(cuò)過了什么?這甚至可能嗎?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.
推薦答案
Stephan T. Lavavej 解釋了為什么這在 這個(gè)視頻.基本上，問題在于編譯器試圖從 both std::vector 和 推導(dǎo)出 BaseTcode>std::function 參數(shù).C++ 中的 lambda 不是 std::function 類型，它是一種未命名的、唯一的非聯(lián)合類型，如果它沒有捕獲列表(空 []).另一方面，可以從任何可能類型的可調(diào)用實(shí)體(函數(shù)指針、成員函數(shù)指針、函數(shù)對象)創(chuàng)建 std::function 對象.
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).
請注意，我個(gè)人不明白為什么您要將傳入的函子限制為該特定簽名(除了通過多態(tài)函數(shù)包裝器間接進(jìn)行的事實(shí)之外，例如 std::function, 遠(yuǎn)比直接調(diào)用函子(甚至可能被內(nèi)聯(lián))低效)，但這是一個(gè)工作版本.基本上，它禁用了 std::function 部分的參數(shù)推導(dǎo)，并且只從 std::vector 參數(shù)推導(dǎo)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 上的實(shí)例.
另一種可能的方法是不直接限制函子類型，而是通過 SFINAE 間接限制:
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 上的實(shí)例.
如果 fun 不接受 T& 類型的參數(shù)或者返回類型不能轉(zhuǎn)換為 ，這個(gè)函數(shù)將從重載集中刪除布爾., void() 使得 f 的返回類型為 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.
這篇關(guān)于Lambda 和 std::function的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！