是否可以使用 boost::filter_iterator 进行输出?

Is it possible to use boost::filter_iterator for output?(是否可以使用 boost::filter_iterator 进行输出?)

问题描述

我使用带有 std::back_inserter 的 std::transform 将元素附加到 std::deque.现在，在某些情况下，转换可能会失败并返回无效对象(比如未初始化的 boost::optional 或空指针).我想从附加中过滤掉无效对象.

I am using std::transform with an std::back_inserter to append elements to an std::deque. Now the transformation may fail and will return a invalid object (say an uninitialized boost::optional or a null pointer) in some cases. I would like to filter out the invalid objects from getting appended.

我考虑过使用 boost::filter_iterator，但不确定如何呈现过滤范围的 end() 参数.

I thought about using boost::filter_iterator, but not sure how to present the end() parameter of the filtered range.

boost::filter_iterator 的文档表明输出过滤是可能的.在这种情况下，我是否应该专门为 std::back_insert_iterator 指定 operator == 以始终返回 false?

The documentation of boost::filter_iterator suggests that output filtering is possible. Should I just specialize operator == for std::back_insert_iterator in this case to always return false?

除此之外，如果我想附加初始化的 boost::optional 或指针的值，我可以链接 boost::filter_iterator 和 boost::indirect_iterator?

In addition to this, if I want to append values of initialized boost::optional or pointers, can I chain boost::filter_iterator and boost::indirect_iterator?

我试图避免推出我自己的 transform_valid 函数，该函数采用可选的 extractor 函数.

I am trying to avoid rolling out my own transform_valid function that takes an optional extractor function.

是否可以使用 filter_iterator 作为输出迭代器?

Is it even possible to use filter_iterator as an output iterator?

推荐答案

为了方便使用，我建议使用 boost range(算法和适配器)，你会写:

I suggest using boost range (algorithms & adaptors) for ease of use, you'd write:

boost::copy(
    data | transformed(makeT) | filtered(validate) /* | indirected */, 
    std::back_inserter(queue));

这是一个完整的工作示例:

Here is a complete working example of that:

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

#include <vector>
#include <deque>

typedef boost::optional<int> T;
typedef std::deque<T> Q;

static T makeT(int i)
{
    if (i%2) return T();
    else     return i;
}

static bool validate(const T& optional) 
{ 
    return (bool) optional; // select the optional that had a value set
}

int main()
{
    static const int data[] =  { 1,2,3,4,5,6,7,8,9 };

    Q q;

    using boost::adaptors::filtered;
    using boost::adaptors::transformed;

    // note how Boost Range elegantly supports an int[] as an input range
    boost::copy(data | transformed(makeT) | filtered(validate), std::back_inserter(q));

    // demo output: 2, 4, 6, 8 printed
    for (Q::const_iterator it=q.begin(); it!=q.end(); ++it)
    {
        std::cout << (*it? "set" : "unset") << "	" << it->get_value_or(0) << std::endl;
    }

    return 0;
}

更新

从这个答案中得到一点帮助:使用 boost::optional 和 boost::adaptors::间接

With a little help from this answer: Use boost::optional together with boost::adaptors::indirected

我现在包含了一个使用 indirected 范围适配器以及队列的直接输出的优雅演示(取消引用选项):

I now include an elegant demonstration of using the indirected range adaptor as well for immediate output of the queue (dereferencing the optionals):

_{请注意，对于(智能)指针类型，显然不需要提供 pointee<> 特化.我认为这是设计使然:可选<>不是，也不建模，指针}

_{Note that for (smart) pointer types there would obviously be no need to provide the pointee<> specialisation. I reckon this is by design: optional<> is not, and does not model, a pointer}

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

#include <boost/optional.hpp>

namespace boost {
    template<typename P> struct pointee<optional<P> > {
        typedef typename optional<P>::value_type type;
    };
}

typedef boost::optional<int> T;

static T    makeT(int i)                { return i%2?  T() : i; }
static bool validate(const T& optional) { return (bool) optional; }

int main() {
    using namespace boost::adaptors;

    static int data[] =  { 1,2,3,4,5,6,7,8,9 };
    boost::copy(data | transformed(makeT) 
                     | filtered(validate) 
                     | indirected, 
                     std::ostream_iterator<int>(std::cout, ", "));
}

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