升压转换迭代器和C ++ 11的lambda [英] boost transform iterator and c++11 lambda

问题描述

我想使用boost ::适配器::通过向适配器提供的C ++ 0x的lambda转变。

I'm trying to use boost::adaptors::transformed by providing a c++0x lambda to the adaptor.

以下code不能编译。我使用的是G ++ 4.6.2提升1.48。

The following code does not compile. I'm using g++ 4.6.2 with boost 1.48.

#include <iostream>
#include <vector>

#include <boost/range/adaptors.hpp>
#include <boost/range/algorithm.hpp>

using namespace std;
namespace br    = boost::range;
namespace badpt = boost::adaptors;


int main()
{  
  vector<int> a = {0,3,1,};
  vector<int> b = {100,200,300,400};

  auto my_ftor = [&b](int r)->int{return b[r];};

  cout<<*br::max_element(a|badpt::transformed(my_ftor))<<endl;
}

这是我做错了什么在这里的任何想法？

Any ideas on what I'm doing wrong here?

推荐答案

好lambda表达式没有发挥好，因为它们没有缺省构造，这是必要的迭代器。这是一个包装我使用的lambda表达式：

Well lambdas don't play nice, since they are not default constructible, which is necessary for iterators. Here is a wrapper I use for lambdas:

#define RETURNS(...) -> decltype(__VA_ARGS__) { return (__VA_ARGS__); }

template<class Fun>
struct function_object
{
    boost::optional<Fun> f;

    function_object()
    {}
    function_object(Fun f): f(f)
    {}

    function_object(const function_object & rhs) : f(rhs.f)
    {}

    // Assignment operator is just a copy construction, which does not provide
    // the strong exception guarantee.
    function_object& operator=(const function_object& rhs)
    {
        if (this != &rhs)
        {
            this->~function_object();
            new (this) function_object(rhs);
        }
        return *this;
    }

    template<class F>
    struct result
    {};

    template<class F, class T>
    struct result<F(T)>
    {
        typedef decltype(std::declval<Fun>()(std::declval<T>())) type;
    };

    template<class T>
    auto operator()(T && x) const RETURNS((*f)(std::forward<T>(x)))

    template<class T>
    auto operator()(T && x) RETURNS((*f)(std::forward<T>(x)))
};

template<class F>
function_object<F> make_function_object(F f)
{
    return function_object<F>(f);
}

然后，你可以这样做：

Then you can just do this:

int main()
{  
  vector<int> a = {0,3,1,};
  vector<int> b = {100,200,300,400};

  cout<<*br::max_element(a|badpt::transformed(make_function_object([&b](int r)->int{return b[r];};)))<<endl;
}

这篇关于升压转换迭代器和C ++ 11的lambda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IT屋！