当传递容器参数时,range-v3中的
view::zip
创建一个由对原始元素的引用组成的元组视图。将zipped视图传递给
sort
可以就地对元素进行排序。即,以下程序:
#include <vector>
#include <string>
#include <iostream>
#include <range/v3/algorithm.hpp>
#include <range/v3/view.hpp>
using namespace ranges;
template <std::size_t N>
struct get_n {
template <typename T>
auto operator()(T&& t) const ->
decltype(std::get<N>(std::forward<T>(t))) {
return std::get<N>(std::forward<T>(t));
}
};
namespace ranges {
template <class T, class U>
std::ostream& operator << (std::ostream& os, common_pair<T, U> const& p) {
return os << '(' << p.first << ", " << p.second << ')';
}
}
int main() {
std::vector<std::string> names {"john", "bob", "alice"};
std::vector<int> ages {32, 19, 35};
auto zipped = view::zip(names, ages);
std::cout << "Before: Names: " << view::all(names) << '\n'
<< " Ages: " << view::all(ages) << '\n'
<< " Zipped: " << zipped << '\n';
sort(zipped, less{}, get_n<1>{});
std::cout << " After: Names: " << view::all(names) << '\n'
<< " Ages: " << view::all(ages) << '\n'
<< " Zipped: " << zipped << '\n';
}
输出结果:
原始数据: 姓名:[john,bob,alice]
年龄:[32,19,35]
合并后: [(john, 32),(bob, 19),(alice, 35)]
现在数据: 姓名:[bob,john,alice]
年龄:[19,32,35]
合并后: [(bob, 19),(john, 32),(alice, 35)]
Coliru上的实时示例.