是否有从容器转换的标准方法<Type1>到容器<Type2>?
我有两个类A和B，并且存在一个隐式转换运算符从一个到另一个，所以:
I have two classes A and B, and an implicit conversion operator exists to go from one to the other, so that:
A a;
B b;
b = a; // Works
是否有将 std::list<A> 转换为 std::list<B> 的标准方法?(甚至从 std::vector<A> 到 std::list<B>).
Is there a standard way to convert a std::list<A> to a std::list<B> ? (Or even from std::vector<A> to a std::list<B>).
我知道我可以遍历列表并逐项构建第二个列表，但我想知道是否有更优雅的解决方案.
I know I can iterate trough to the list and build the second list item by item, but I wonder if there is a more elegant solution.
不幸的是，我不能使用 boost，但出于好奇，如果 boost 可以处理这个问题，我也很乐意知道如何处理.
Unfortunately I cannot use boost but out of curiosity as a bonus question, if boost can handle this, I'd be happy to know how too.
推荐答案
嗯，是的.每个序列容器类型都有一个模板构造函数，它接受一对迭代器(一个迭代器范围)作为输入.它可以用于从另一个序列构造一个序列，而不管序列类型如何，只要序列元素类型可以相互转换.比如说
Well, yes. Each sequence container type has a template constructor that takes a pair of iterators (an iterator range) as an input. It can be used to construct one sequence from another, regardless of the sequence types, as long as the sequence element types are convertible to each other. Like for example
std::vector<A> v;
...
std::list<B> l(v.begin(), v.end());
序列容器也有 assign 成员函数，它对赋值语义(与初始化语义相反)做同样的事情.
Also sequence containers have assign member function which does the same thing with assignment semantics (as opposed to initialization semantics).
std::vector<A> v;
std::list<B> l;
...
l.assign(v.begin(), v.end()); // replaces the contents of `l`
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