如何使用boost :: adaptors :: transformed从模板类和向量生成范围? [英] How do I use boost::adaptors::transformed to produce a range from a templated class and a vector?

问题描述

我问了一个有关使用lambda早期实现类似功能的问题，但

I asked a question about using a lambda to achieve something similar earlier but wasn't able to get this working so I've tried approach the problem using a functor instead. This is probably neater anyway in the sense that it doesn't involve constructing std::function objects and is closer to the example case set out in the documentation.

这是一个简单的设置，它说明了我的问题:

Here is a simple setup that illustrates my problem:

#include <boost/range/adaptor/transformed.hpp>
#include <vector>

// some structure
template <typename T>
struct MyStruct
{
    MyStruct(T t)
    {
    }
};

template <typename T>
struct Converter
{
    using return_type = MyStruct<T>;

    return_type operator()(const T& value)
    {
        return return_type(value);
    }
};

int main(int argc, const char* argv[])
{
    std::vector<int> vec {1, 2, 3};

    auto val = vec | boost::adaptors::transformed(Converter<int>());

    return 0;
}

当我尝试对此进行编译时，出现以下错误消息:

When I try to compile this I am getting the following error message:

/home/user/packages/boost/mpl/eval_if.hpp:38:31:错误:未命名类型 "boost::mpl::eval_if<boost::is_same<boost::use_default, boost::use_default>, boost::result_of<const Converter<int>(int&)>, boost::mpl::identity<boost::use_default> >::f_ {aka struct boost::result_of<const Converter<int>(int&)>}"

/home/user/packages/boost/mpl/eval_if.hpp:38:31: error: no type named ‘type’ in ‘boost::mpl::eval_if<boost::is_same<boost::use_default, boost::use_default>, boost::result_of<const Converter<int>(int&)>, boost::mpl::identity<boost::use_default> >::f_ {aka struct boost::result_of<const Converter<int>(int&)>}’

我不确定该怎么做.我无法在代码中发现任何错误.有什么想法吗?

I'm not sure what to make of this. I can't spot any errors in the code. Any ideas?

推荐答案

该错误告诉您boost::result_of<const Converter<int>(int&)>没有type成员.换句话说，使用const Converter<int>对象时，函数调用运算符不起作用.一旦知道问题所在，就很容易发现问题所在:

The error tells you boost::result_of<const Converter<int>(int&)> doesn't have the type member. In other words, the function call operator doesn't work when a const Converter<int> object is used. Once you know the problem, it's easy to see what's wrong:

return_type operator()(const T& value) const
//                                     ^^^^^
{
    return return_type(value);
}

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