程序目标:输入一个字符串,竖向输出该字符串。使用string和动态分配内存机制。代码如下:

#include<iostream>
#include "stdafx.h"
#include<cstring>
int main()
{
using namespace std;
string s= cin.getline;
char * a = new char[sizeof(s)];
for (int m = 0;m < sizeof(s);m++)
a[m] = s[m];
for (int n = 0;n < sizeof(s);n++)
cout << a[n] << endl;
delete[] a;
cin.get();
cin.get();
return 0;
}

编译器:Visual Studio 2015

显示编译错误,但是不知道怎么修改。先记录在这。希望大神不吝赐教。

错误信息:

严重性 代码 说明 项目 文件 行 禁止显示状态
错误 C2146 语法错误: 缺少“;”(在标识符“s”的前面) MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 10
错误 C2065 “s”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 10
错误 C2065 “cin”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 10
错误 C2228 “.getline”的左边必须有类/结构/联合 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 10
错误 C2065 “s”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 11
错误 C2065 “s”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 12
错误 C2065 “s”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 13
错误 C2065 “s”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 14
错误 C2065 “cout”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 15
错误 C2065 “endl”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 15
错误 C2065 “cin”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 17
错误 C2228 “.get”的左边必须有类/结构/联合 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 17
错误 C2065 “cin”: 未声明的标识符 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 18
错误 C2228 “.get”的左边必须有类/结构/联合 MyProject_001 c:\users\administrator\documents\visual studio 2015\projects\myproject_01\myproject_001\myproject_001\myproject_001.cpp 18

####希望用一个错误来开始我的博客园之旅,只有不停的纠错,才能提高####

2018/3/2晚11点30分写的程序(C++)的更多相关文章

  1. 大一C语言学习笔记(11)---编程篇--写一个程序,可以获取从键盘上输入的的三个数,并能够判断是否可以以这三个数字作为边长来构成一个三角形,如果可以的话,输出此三角形的周长及面积,要求 0 bug;

    考核内容: 写一个程序,可以获取从键盘上输入的的三个数,并能够判断是否可以以这三个数字作为边长来构成一个三角形,如果可以的话,输出此三角形的周长及面积: 答案: #include<stdio.h ...

  2. PTA 07-图5 Saving James Bond - Hard Version (30分)

    07-图5 Saving James Bond - Hard Version   (30分) This time let us consider the situation in the movie ...

  3. PAT A1127 ZigZagging on a Tree (30 分)——二叉树,建树,层序遍历

    Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can ...

  4. 1127 ZigZagging on a Tree (30 分)

    1127 ZigZagging on a Tree (30 分) Suppose that all the keys in a binary tree are distinct positive in ...

  5. 【PAT】1053 Path of Equal Weight(30 分)

    1053 Path of Equal Weight(30 分) Given a non-empty tree with root R, and with weight W​i​​ assigned t ...

  6. 1053 Path of Equal Weight (30 分)

    Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weig ...

  7. pta5-9 Huffman Codes (30分)

    5-9 Huffman Codes   (30分) In 1953, David A. Huffman published his paper "A Method for the Const ...

  8. A1095 Cars on Campus (30)(30 分)

    A1095 Cars on Campus (30)(30 分) Zhejiang University has 6 campuses and a lot of gates. From each gat ...

  9. PTA 11-散列4 Hard Version (30分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/680 5-18 Hashing - Hard Version   (30分) Given ...

随机推荐

  1. 通过Azure 存储账号URL鉴别是标准磁盘还是高性能磁盘

    对于不知道虚拟机磁盘是标准磁盘还是高性能磁盘时,我们可以通过nslookup解析存储账号的URL,来判断存储账号的类型,从而得知虚拟磁盘的类型 1.标准存储账号的解析结果,字母"st&quo ...

  2. junit测试套件

    在实际项目中,随着项目进度的开展,单元测试类会越来越多,可是直到现在我们还只会一个一个的单独运行测试类,这在实际项目实践中肯定是不可行的.为了解决这个问题,JUnit 提供了一种批量运行测试类的方法, ...

  3. 输入和输出--javase中的路径

    就目前为止, javase中经常用到路径来读取一个资源文件的所有情况都已经整理在博客里面了,这里做一个统一的整理: 1,IO流来读取一个文件,比如说new FileInputStream(" ...

  4. @interface注解类、 @Target:注解的作用目标 @Retention

    public @interface xxx 定义注解 @interface 不是interface,是注解类 是jdk1.5之后加入的,java没有给它新的关键字,所以就用@interface 这么个 ...

  5. 我的运维之旅-查找文本的linux命令

    小伙伴们肯定都遇到这么尴尬场景,线上服务出问题了,老大一直在问什么问题导致的,而你由于对查找文本的命令不太熟,鼓捣了半天才找到那条 异常日志,而这时可能半个小时都已经过去了.老大可能对你失望透顶了.讲 ...

  6. 【转】EI收录的中国期刊

    ISSN     期刊名  0567-7718 Acta Mechanica Sinica  1006-7191 Acta Metallurgica Sinica (English Letters)  ...

  7. TCP/IP协议全解析 三次握手与四次挥手[转]

    所谓三次握手(Three-Way Handshake)即建立TCP连接,就是指建立一个TCP连接时,需要客户端和服务端总共发送3个包以确认连接的建立.所谓四次挥手(Four-Way Wavehand) ...

  8. 封装的应用【example_Array工具】

    定义一个数组工具[ArrayTool]封装其方法,ArrayDemo调用数组工具ArrayTool package new_Object; //封装多个个功能 class ArrayTool{ //1 ...

  9. 123 A. Prime Permutation

    链接 http://codeforces.com/contest/123/problem/A 题目 You are given a string s, consisting of small Lati ...

  10. toString()方法细节

    toString(),每一个非基本类型的对象都有一个toString()方法,当编译器需要一个Sting,而你只有一个对象时,该方法会自动调用. class WaterSource { private ...