重写对集合的访问以避免“双重"访问.发现

本文介绍了重写对集合的访问以避免“双重"访问.发现的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有这样的代码:

std::unordered_map<int64_t /*id_ord*/, LimitOrder> futOrders; auto i = futOrders.find(orderId); if (i == futOrders.end()) { LimitOrder& newOrder = futOrders[orderId]; // work } else { LimitOrder& futOrder = i->second; // another work }

我在这里执行两次查找":第一次: auto i = futOrders.find(orderId); 第二次: LimitOrder&newOrder = futOrders [orderId];

Here I execute "find" twice: first time: auto i = futOrders.find(orderId); second time: LimitOrder& newOrder = futOrders[orderId];

我可以以某种方式重写它以避免重复查找"吗?

Can i rewrite it somehow to avoid "double find"?

推荐答案

您可以执行 emplace ，并检查返回值以了解是否插入了该项目:

You can perform an emplace, and check the return value to know whether the item was inserted or not:

std::unordered_map<int64_t /*id_ord*/, LimitOrder> futOrders; auto i = futOrders.emplace( std::piecewise_construct, std::tie(orderId), std::make_tuple()); if (i.second) { LimitOrder& newOrder = i.first->second; // work } else { LimitOrder& futOrder = i.first->second; // another work }