Question

If I m constructing a non-trivial container like a custom map for example, and want to share code between the emplace and insert functions, one approach would be to have the insert function simply call emplace with the supplied element as the argument. eg.:

template<typename... args>
iterator emplace(int id, args &&... params)
{
    // Do complex stuff here, get a location
    *location = value_type(std::forward<args>(params) ...);
}


iterator insert(int id, const value_type &element)
{
    emplace(id, element);
}

Which in the case of insert & should simply call the copy constructor if I understand correctly? However in the case of insert && it becomes a little more confusing:

iterator insert(int id, value_type &&element)
{
    emplace(id, std::move(element));
}

<代码>std:move a. 折中元件,罚款,但将<代码>td:forward/code> 如何适当处理? e. 非属地的<代码> 数值_ 类型与非违约移动建筑商的,之上的<代码>emplace建筑线将正确地称作移动建筑商?

Answer 1

在<代码>insert &中,如果我正确理解,究竟应当把复印件放在什么位置?

Yes, you re giving std::forward a const lvalue in this scenario, and std::forward yields a const lvalue. With that, the copy constructor is called.

<代码>std:move a. 折中元件,罚款,但将<代码>td:forward/code> 如何适当处理?

Yes. std::forward turns rvalues into xvalues and lvalues into lvalues. As long as you don t hand it an lvalue, it won t give you an lvalue. This subsequently means that the move constructor will be called, as intended.

In args &&... params, args deduces to value_type, making params an rvalue reference when the function template is instantiated. The second overload of std::forward is called, and std::forward<args>(params) yields an xvalue.

It s actually quite common to implement push_back or insert for containers in terms of emplace_back or emplace.

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