c++ primer（第五版）学习笔记及习题答案代码版（第一章）

笔记较为零散，都是自己不熟悉的知识点。
习题答案至于一个.cc中，需要运行某一题直接修改#define NUM**， 如运行第一题为#define NUM11，题1.24定义为NUM124
chapter 1

1、std::cout << "Entertwo numbers:" << std::endl;等价于

std::cout<< "Enter two numbers:";

std::cout << std::endl;

2、命名空间的作用是防止与库中定义的名字发生冲突，其副作用是，当使用标准库中的名字时必须显示地表达出使用的是命名空间std下的名字。

3、for (int val = 1; val <= 10; ++val)

对于for循环内的val变量，当结束循环后不再可访问

4、文件结束符

ctrl+d for unix/linux/ etc 
ctrl+z for windows/dos

一旦测试失败，while 终止并退出循环体，执行 while 之后的语句。该语

句在输出 sum 后输出 endl，endl 输出换行并刷新与 cout 相关联的缓冲区。

最后，执行 return，通常返回零表示程序成功运行完毕。

5、 exit(-1)或者return(-1)为什么shell得到的退出码是255？

http://www.cnblogs.com/tangdoudou/p/3385149.html

exit或return，只能使用0~255之间的值，-1的unsigned值就是255.

#include<iostream>#include "Sales_item.h"#define NUM120//using namespace std;int answerEample(int value1,int value2);int main(){/*1.1*/#ifdef NUM11        cout<<"C++程序一般涉及两类文件：头文件和源文件。文件后缀通常表明文件的类型,"                                "如头文件后缀可以是.h或.hpp;源文件的后缀可以是.cc或.cpp，具体的后缀与使用的编译器有关"<<endl;#endif/*1.2*/#ifdef NUM12        cout<<"window和linux系统下不会报告main函数的运行失败，所以程序返回-1或返回1在运行效果上并没有什么不同,"                        "但是,DOS命令下运行程序，然后键入echo %ERRORLEVEL%,则会显示返回值-1，linux系统下，键入echo $? 则会返回255"<<endl;#endif /*1.3*/#ifdef NUM13        cout<<"Hello world"<<endl;#endif /*1.4*/#ifdef NUM14        cout<<"Enter two number: "<<endl;        int v1, v2;        cin >> v1 >> v2;        cout<< "The product of " <<v1<<" and "<<v2<<" is "<<v1*v2<<endl;#endif /*1.5*/#ifdef NUM15        cout<<"Enter two number: "<<endl;        int v1, v2;        cin >> v1 >> v2;        cout<< "The product of ";        cout<<v1;        cout<<" and ";        cout<<v2;        cout<<" is ";        cout<<v1*v2;        cout<<endl;#endif /*1.6*/#ifdef NUM16        cout<<"代码不合法"                "第1,2,3行的末尾有分号，表示这段代码分别构成三条语句."                "'<<'是二元操作符，在第2,3两行中,第一个'<<'缺少左操作数，因此不合法。应该在2,3行开头加上cout. "<<endl;#endif/*1.7*/#ifdef NUM17/* * commet pairs /(两个**) /cannot nest * "cannot nest" is considered source code. * as is the rest of the problem */        cout<<"错误信息：注释不可嵌套。"<<endl;#endif/*1.8*/#ifdef NUM18        cout<< "第1,2,4行合法。"                "第3行<<操作符之后到第二个双引号之前的部分被注释掉了，导致<<操作符的右操作数不是一个完整的字符串."<<endl;#endif/*1.9*/#ifdef NUM19        int sum(0), val(50);        while(val <= 100){                sum += val;                ++val;        }        cout<<" Sum of 50 to 100 is: "<<sum<<endl;#endif/*1.10*/#ifdef NUM110        int i(10);        while(i >= 0){                cout<< i << " ";                --i;        }        cout<<endl;#endif/*1.11*/#ifdef NUM111        int v1,v2,low,high;        cout<< "Enter two numbers: "<<endl;        cin >> v1 >> v2;        while(v1 <= v2){                cout<< v1 <<" ";                ++v1;        }#endif/*1.12*/#ifdef NUM112        int sum = 0;        for(int i = -100; i<=100; ++i)                sum += i;        cout<<"-100 到 100 整数相加: "<<sum<<endl;#endif/*1.13*/#ifdef NUM113        int sum(0), val(50);        for(val=50; val <= 100; ++val)                sum += val;                cout<< "The sum of numbers from 50 to 100: "<<sum <<endl;        int i(10);        for(i=10; i>=0; --i)                cout<< i <<" ";        cout<<endl;        int v1,v2,low,high;        cout<< "Enter two numbers: "<<endl;        cin >> v1 >> v2;        if(v1 <= v2){                low  = v1;                high = v2;        }else{                low = v2;                high = v1;        }        for(;v1 <= v2; ++v1)                cout<<v1 <<" ";        cout<<endl;#endif/*1.14*/#ifdef NUM114        cout<<" for循环,循环控制变量的初始化和修改都放在语句头部分，形式简洁，特别适用于循环次数已知的情况，"                " while循环,循环控制变量的初始化一般放在while循环的之前，循环控制变量的修改一般放在循环体中，"                "形式不如for简洁，但是比较适用于循环次数不易预知的情况下。"<<endl;#endif/*1.16*/#ifdef NUM116        int sum(0), val;        while(cin >> val)                sum += val;        cout<< "The sum is: "<<sum <<endl;#endif/*1.17*/#ifdef NUM117        cout<<"下载我上传资源中的第五版源代码测试。"<<endl;#endif/*1.19*/#ifdef NUM119        int v1,v2,low,high;        cout<< "Enter two numbers: "<<endl;        cin >> v1 >> v2;        if(v1 <= v2){                low  = v1;                high = v2;        }else{                low = v2;                high = v1;        }        while(v1 <= v2){                cout<< v1 <<" ";                ++v1;        }#endif/*1.20*/#ifdef NUM120        Sales_item book;        cout<<"Enter transactions: "<<endl;        while(cin>> book){                cout << "ISBN, number of copies sold, "                        " total revenue, and average price are: "<<endl;                cout<< book <<endl;        }#endif/*1.21*/#ifdef NUM121        Sales_item book1, book2;        cin >> book1 >> book2;        if(book1.same_isbn(book2))                cout<<book1 + book2 <<endl;        else                cout<<"The two books have different ISBN "<<endl;#endif/*1.22*/#ifdef NUM122        Sales_item total, book;        cout<<" Enter transactions: "<<endl;        if(cin>>total){                while(cin >> book){                        if(total.same_isbn(book))                                total = total + book;                        else                                cout<<"differ ISNB "<<endl;                        return -1;                }                cout << "ISBN, number of copies sold, "                        " total revenue, and average price are: "<<endl;                cout<<total<<endl;        }else                cout<<"Input error " <<endl;                return -255;#endif/*1.23 */#ifdef NUM123        Sales_item book1,book2;        int count(0);        cout<<"Enter transctions: "<<endl;        cin >> book1;        count = 1;        while(cin >> book2)                if(book1.same_isbn(book2))                        ++count;                else {                        cout<<"Transaction amount of previous ISBN: "<<endl;                        book1 = book2;                        count = 1;                }        cout << "Transction amount of the last ISBN "<< count  <<endl;#endif

参考资料：
c++ primer中文版第五版，电子工业出版社。
c++ primer第四版习题解答，人民邮电出版社。