why std::stack has separate top() and pop()
SGI explanation: http://www.sgi.com/tech/stl/stack.html One might wonder why pop() returns void, instead of value_type. That is, why must one use top() and pop() to examine and remove the top element, instead of combining the two in a single member function? In fact, there is a good reason for this design. If pop() returned the top element, it would have to return by value rather than by reference: return by reference would create a dangling pointer. Return by value, however, is inefficient: it involves at least one redundant copy constructor call. Since it is impossible for pop() to return a value in such a way as to be both efficient and correct, it is more sensible for it to return no value at all and to require clients to use top() to inspect the value at the top of the stack.
std::stack < T > is a template. If pop() returned the top element, it would have to return by value rather than by reference as per the of above explanation. That means, at the caller side it must be copied in an another T type of object. That involves a copy constructor or copy assignment operator call. What if this type T is sophisticated enough and it throws an exception during copy construction or copy assignment? In that case, the rvalue, i.e. the stack top (returned by value) is simply lost and there is no other way to retrieve it from the stack as the stack's pop operation is successfully completed!
the 2nd point might not be the reason of the initial design as STL is introduced into C++ before exceptions; it's a good point to make though.
why std::stack has separate top() and pop()的更多相关文章
- C++ std::stack
std::stack template <class T, class Container = deque<T> > class stack; LIFO stack Stack ...
- C++ std::stack 基本用法
#include <iostream> #include <string> #include <stack> // https://zh.cppreference. ...
- 从deque到std::stack,std::queue,再到iOS 中NSArray(CFArray)
从deque到std::stack,std::queue,再到iOS 中NSArray(CFArray) deque deque双端队列,分段连续空间数据结构,由中控的map(与其说map,不如说是数 ...
- 设计一个Stack,要求Push、Pop、获取最大最小值时间复杂度都为O(1)
面试的时候,面试官让设计一个栈,要求有Push.Pop和获取最大最小值的操作,并且所有的操作都能够在O(1)的时间复杂度完成. 当时真没啥思路,后来在网上查了一下,恍然大悟,只能恨自己见识短浅.思路不 ...
- 华为笔试题--LISP括号匹配 解析及源码实现
在17年校招中3道题目AC却无缘华为面试,大概是华为和东华互不待见吧!分享一道华为笔试原题,共同进步! ************************************************ ...
- C++并发编程学习笔记
// // main.cpp // test1 // // Created by sofard on 2018/12/27. // Copyright © 2018年 dapshen. All ...
- C++并发编程 02 数据共享
在<C++并发编程实战>这本书中第3章主要将的是多线程之间的数据共享同步问题.在多线程之间需要进行数据同步的主要是条件竞争. 1 std::lock_guard<std::mute ...
- [POJ2337]Catenyms
题目大意: 定义一个catenym是一对单词,满足第一个单词的末尾字符与第二个单词的开头字符相等. 定义复合catenym是一些单词,满足第i个单词的末尾字符与第i+1个单词的开头字符相等. 给你n个 ...
- 北京清北 综合强化班 Day3
括号序列(bracket) Time Limit:1000ms Memory Limit:128MB 题目描述 LYK有一个括号序列,但这个序列不一定合法. 一个合法的括号序列如下: ()是合法的 ...
随机推荐
- Centos7重新安装yum
Centos7重新安装yum rpm -qa|grep yum 然后用下面的命令删除出现的xxx包: rpm -e --nodeps xxx 下载 python-urlgrabber-3.10-8.e ...
- An Overview of Query Optimization in Relational Systems
An Overview of Query Optimization in Relational Systems
- samba服务器的搭建和配置
案例: 公司有两个部门, sales / market . 分别有成员 jack / tom 和 zhang / shen . 公司需求是这样的, 本部门资料禁止其他部门访问, 本部门成员之间不能干扰 ...
- mysq'l系列之10.mysql优化&权限控制
网站打开慢如何排查 1.打开网页, 用谷歌浏览器F12, 查看network: 哪个加载时间长就优化哪个 2.如果是数据库问题 2.1 查看大体情况 # top # uptime //load av ...
- Git with SVN
1)GIT是分布式的,SVN不是: 这 是GIT和其它非分布式的版本控制系统,例如SVN,CVS等,最核心的区别.好处是跟其他同事不会有太多的冲突,自己写的代码放在自己电脑上,一段时间后再提交.合并, ...
- 我的Java开发学习之旅------>工具类:Java使用正则表达式分离出字符串中的中文和英文
今天看到一个工具类使用正则表达式将一大段字符串中的中文和英文都分离出来了,在此记录一下,读者可以收藏! import java.util.ArrayList; import java.util.Col ...
- mysql 二:操作表
的存储.在操作表之前,首先要用选定数据库,因为表都是建立在对应的数据库里面的.在这里我们使用之前建立的test数据库 mysql> use test; Database changed 创建表的 ...
- POJ3984 迷宫问题【水BFS】
版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u011775691/article/details/28050277 #include <cs ...
- PHP中的排序函数sort、asort、rsort、krsort、ksort区别分析(转)
sort() 函数用于对数组单元从低到高进行排序. rsort() 函数用于对数组单元从高到低进行排序. asort() 函数用于对数组单元从低到高进行排序并保持索引关系. arsort() 函数用于 ...
- 04-树7 二叉搜索树的操作集(30 point(s)) 【Tree】
04-树7 二叉搜索树的操作集(30 point(s)) 本题要求实现给定二叉搜索树的5种常用操作. 函数接口定义: BinTree Insert( BinTree BST, ElementType ...