什么是同时迭代两个或更多容器的最好方法 [英] What's the best way to iterate over two or more containers simultaneously
问题描述
C ++ 11提供了多种方法来遍历容器。例如:
C++11 provides multiple ways to iterate over containers. For example:
for(auto c : container) fun(c)
std :: for_each
std::for_each
for_each(container.begin(),container.end(),fun)
但是,建议使用同样大小的两个(或更多)容器来完成类似的操作:
However what is the recommended way to iterate over two (or more) containers of the same size to accomplish something like:
for(unsigned i = 0; i < containerA.size(); ++i) {
containerA[i] = containerB[i];
}
推荐答案
使用
std::copy_n(contB.begin(), contA.size(), contA.begin())
对于更一般的情况,可以使用Boost.Iterator的 zip_iterator
,有一个小功能,使其可用于基于范围的for循环。在大多数情况下,这将工作:
For the more general case, you can use Boost.Iterator's zip_iterator
, with a small function to make it usable in range-based for loops. For most cases, this will work:
template<class... Conts>
auto zip_range(Conts&... conts)
-> decltype(boost::make_iterator_range(
boost::make_zip_iterator(boost::make_tuple(conts.begin()...)),
boost::make_zip_iterator(boost::make_tuple(conts.end()...))))
{
return {boost::make_zip_iterator(boost::make_tuple(conts.begin()...)),
boost::make_zip_iterator(boost::make_tuple(conts.end()...))};
}
// ...
for(auto&& t : zip_range(contA, contB))
std::cout << t.get<0>() << " : " << t.get<1>() << "\n";
但是,对于完整的一般性,你可能想要更像 this ,对于没有成员 begin()
/ end()的数组和用户定义类型,
但 do 在其命名空间中有 begin
/ 。此外,这将允许用户通过
zip_c ...
函数专门获取 const
。
However, for full-blown genericity, you probably want something more like this, which will work correctly for arrays and user-defined types that don't have member begin()
/end()
but do have begin
/end
functions in their namespace. Also, this will allow the user to specifically get const
access through the zip_c...
functions.
如果你是一个好主意的错误信息,像我一样,那么你可能想要这个,它检查是否有任何临时容器传递给任何 zip _...
函数,并打印一个不错的错误消息。
And if you're an advocate of nice error messages, like me, then you probably want this, which checks if any temporary containers were passed to any of the zip_...
functions, and prints a nice error message if so.
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