自定义范围为boost :: range的库 [英] Custom range for boost::range library
问题描述
我正在使用boost :: range库编写过滤器和地图算法:
I’m writing filter and map algorithms using boost::range library:
template <class Range> struct Converter
{
Converter(const Range& p_range) : m_range(p_range) {}
template<class OutContainer> operator OutContainer() const
{
return {m_range.begin(), m_range.end()};
}
private:
Range m_range;
};
template<class Range> Converter<Range> convert(const Range& p_range) { return {p_range}; }
template<class Range, class Fun> auto map(Range&& p_range, Fun&& p_fun)
{
return convert(p_range | boost::adaptors::transformed(p_fun));
}
template<class Range, class Pred> auto filter(Range&& p_range, Pred&& p_pred)
{
return convert(p_range | boost::adaptors::filtered(p_pred));
}
现在我可以像这样使用它们:
Right now I can use them like this:
std::vector<int> l_in = {1, 2, 3, 4, 5};
std::vector<int> l_tmp_out = filter(l_in, [](int p){ return p < 4; });
std::vector<int> l_out = map(l_tmp_out, [](int p){ return p + 5; });
我也想这样写代码:
map(filter(l_in, [](int p){ return p < 4; }), [](int p){ return p + 5; });
不幸的是,我的Converter类没有使用boost :: range算法组成,因此该示例无法编译.我正在寻找一种改变它的正确方法.
Unfortunately my Converter class does not compose with boost::range algorithms so this example does not compile. I'm looking for a proper way to change that.
更新
我跟随了@sehe链接,结果发现我要做的就是将这四行添加到Converter类:
I followed @sehe link and it turned out that all I had to do was to add this four lines to Converter class:
using iterator = typename Range::iterator;
using const_iterator = typename Range::const_iterator;
auto begin() const { return m_range.begin(); }
auto end() const { return m_range.end(); }
推荐答案
这是我的观点:
#include <boost/range.hpp>
#include <boost/range/adaptors.hpp>
#include <boost/range/algorithm.hpp>
#include <iostream>
#include <vector>
namespace MyRange {
template <typename R> struct Proxy {
Proxy(R&& r) : _r(std::move(r)) {}
Proxy(R const& r) : _r(r) {}
template <typename OutContainer> operator OutContainer() const {
return boost::copy_range<OutContainer>(_r);
}
using iterator = typename boost::range_mutable_iterator<R>::type;
using const_iterator = typename boost::range_const_iterator<R>::type;
auto begin() const { return range_begin(_r); }
auto end() const { return range_end(_r); }
auto begin() { return range_begin(_r); }
auto end() { return range_end(_r); }
private:
R _r;
};
template <typename R> auto make_proxy(R&& r) { return Proxy<R>(std::forward<R>(r)); }
template <typename Range, typename Fun> auto map(Range&& p_range, Fun&& p_fun) {
return make_proxy(std::forward<Range>(p_range) | boost::adaptors::transformed(std::forward<Fun>(p_fun)));
}
template <typename Range, typename Pred> auto filter(Range&& p_range, Pred&& p_pred) {
return make_proxy(std::forward<Range>(p_range) | boost::adaptors::filtered(std::forward<Pred>(p_pred)));
}
}
int main() {
using namespace MyRange;
{
std::vector<int> l_in = {1, 2, 3, 4, 5};
std::vector<int> l_tmp_out = filter(l_in, [](int p){ return p < 4; });
std::vector<int> l_out = map(l_tmp_out, [](int p){ return p + 5; });
boost::copy(l_out, std::ostream_iterator<int>(std::cout << "\nfirst:\t", "; "));
}
{
boost::copy(
map(
filter(
std::vector<int> { 1,2,3,4,5 },
[](int p){ return p < 4; }),
[](int p){ return p + 5; }),
std::ostream_iterator<int>(std::cout << "\nsecond:\t", "; "));
}
}
打印
first: 6; 7; 8;
second: 6; 7; 8;
注释
- 它更准确地使用
std::forward<>
- 它使用const/non-const迭代器
-
它使用Boost Range特征(
range_mutable_iterator<>
等),而不是假设嵌套typedef进行硬编码.这样一来,事情就可以与其他范围(例如std::array<>
甚至int (&)[]
)一起使用.NOTES
- it uses
std::forward<>
more accurately - it uses const/non-const iterators
it uses Boost Range traits (
range_mutable_iterator<>
etc.) instead of hardcoding assuming nested typedefs. This allows things to work with other ranges (e.g.std::array<>
or evenint (&)[]
).出于类似原因,用户定义的converson运算符使用
boost::copy_range<>
the user-defined converson operator uses
boost::copy_range<>
for similar reasons这篇关于自定义范围为boost :: range的库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!
- it uses