我怎样才能让的std :: find_if和std ::地图共同努力使用一些Boost库？ [英] How can I make std::find_if and std::map work together using some boost library?

问题描述

这个问题是来自<一个灵感href=\"http://stackoverflow.com/questions/7335546/find-the-first-value-greater-than-user-specified-value-from-a-map-container\">another话题这对这样一个问题：

This question is inspired from another topic which poses this question:

查找地图容器中第一个值大于用户指定的值时

Find the first value greater than user specified value from a map container

这可以通过多种方式来解决。一个典型的C ++ 03的解决方案定义了一个专门的函数（或仿函数），并把它传递给的std :: find_if 作为第三个参数。

which can be solved in several ways. A typical C++03 solution defines a dedicated function (or functor) and pass it to std::find_if as third argument.

在C ++ 11，他就会避免定义的专用的函数（或仿函数），而是可以使用的λ为：

In C++11, one can avoid defining a dedicated function (or functor), and can instead make use of lambda as:

auto it = std:: find_if(m.begin(), mp.end(), 
                    [n](const std::pair<std::string, int> & x) -> bool
                       { return x.second > n; }
                   );

这是<一个href=\"http://stackoverflow.com/questions/7335546/find-the-first-value-greater-than-user-specified-value-from-a-map-container/7335623#7335623\">the接受的答案。

我还在寻找一个短暂而凉爽的解决方案。如果它是一个矢量，然后我就学会了一个很酷的解决方案，它利用 Boost.Phoenix 和解决方案变得非常简洁（的 ideone演示）：

I'm still looking for a short and cool solution. If it were a vector, then I just learnt a cool solution which makes use of Boost.Phoenix and the solution becomes very concise (ideone demo):

std::vector<int> v = ...;
auto it = std::find_if(v.begin(), v.end(), arg1 > 4);

下面 ARG1 是的boost ::凤凰:: arg_names 命名空间定义一个仿函数对象，和前pression ARG1方式＆gt; 4 计算为另一个仿函数然后把它传递给的std :: find_if

Here arg1 is a functor object defined in boost::phoenix::arg_names namespace, and the expression arg1>4 evaluates to another functor which then gets passed to std::find_if.

一个快速测试（ ideone ）

std::cout<< (arg1 > 9)(v) << std::endl; //prints 0 if as v > 9 is false, else 1

//or store the functor first and then use it
const auto & f = arg1 > 9;
std::cout<<  f(v) << std::endl; //prints 0 if as v > 9 is false, else 1

---

我的问题是，我想解决这个问题的地图，以类似的方式。有没有这样的解决方案？是这样的：

---

My question is, I want to solve the map problem, in a similar way. Is there any such solution? Something like:

auto it = std::find_if(m.begin(),mp.end(), (???).second > n); //m is std::map

或者

auto it = std::find_if(m.begin(),mp.end(), at<1>(arg1) > n);  //m is std::map

有关它的工作，前pression AT＆LT; 1 GT;（ARG1）GT; 2 必须评估一个仿函数这需要常量的std ::对＆安培; 作为参数。我的直觉告诉我，提振了该解决方案。 ： - ）

For it to work, the expression at<1>(arg1) > 2 has to evaluate to a functor which takes const std::pair & as argument. My gut feelings tells me that boost has this solution. :-)

推荐答案

事实上，Boost.Fusion和Boost.Phoenix有你想要内置的是什么。

Indeed, Boost.Fusion and Boost.Phoenix have exactly what you want built-in.

如果一个包括必要的头适应 STD ::对＆LT;＆GT; 为符合融合序列，那么可以使用Phoenix的的懒惰版本的boost ::融合:: at_c＆LT;＆GT; 访问的std ::对＆LT;＆GT; ::第一个或的std ::对＆LT;＆GT; ::第二个（一定要的#include＆LT;升压/凤凰/ fusion.hpp＆GT; ）。

If one includes the necessary header to adapt std::pair<> as a conforming Fusion sequence, then one can use Phoenix's lazy version of boost::fusion::at_c<> to access std::pair<>::first or std::pair<>::second (be sure to #include <boost/phoenix/fusion.hpp>).

namespace phx = boost::phoenix;
using phx::arg_names::arg1;

auto it = std::find_if(m.begin(), m.end(), phx::at_c<1>(arg1) > n);

---

编辑：展开样品，用VC ++ 2010 SP1 +升压1.47.0测试：

---

Full sample, tested with VC++ 2010 SP1 + Boost 1.47.0:

#include <algorithm>
#include <map>
#include <string>
#include <iostream>
#include <boost/fusion/include/std_pair.hpp>
#include <boost/phoenix/core.hpp>
#include <boost/phoenix/operator.hpp>
#include <boost/phoenix/fusion.hpp>

int main()
{
    namespace phx = boost::phoenix;
    using phx::arg_names::arg1;

    std::map<std::string, int> m;
    m["foo"]    = 1;
    m["bar"]    = 2;
    m["baz"]    = 3;
    m["qux"]    = 4;
    m["quux"]   = 5;
    m["corge"]  = 6;
    m["grault"] = 7;
    m["garply"] = 8;
    m["waldo"]  = 9;
    m["fred"]   = 10;
    m["plugh"]  = 11;
    m["xyzzy"]  = 12;
    m["thud"]   = 13;

    int const n = 6;
    auto it = std::find_if(m.cbegin(), m.cend(), phx::at_c<1>(arg1) > n);
    if (it != m.cend())
        std::cout << it->first << '\n'; // prints "fred"
}

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