• C++入门程序作业2


    程序在Dev-C++5.5.3版本运行

    结构体的使用

    给结构体赋值,打印出结构体中学生姓名,分数,平均分

    #include <iostream>
    #include <cassert>
    #include <algorithm>
    #include <vector>
    #include <string.h>//不用.h的话可能下面的strcpy用不了
    #include <iterator>
    using namespace std;

    int main()
    {
    struct Student //声明结构体类型Student
    {

    char name[20];
    float score1;
    float score2;
    float score3;
    }student1, student2; //定义两个结构体类型Student的变量student1,student2


    strcpy(student1.name,"小杨");
    student1.score1=60;
    student1.score2=70;
    student1.score3=80;

    strcpy(student2.name,"小李");
    student2.score1=96;
    student2.score2=97;
    student2.score3=98;
    int a=0;
    a= (student1.score1+student1.score2+student1.score3)/3;
    cout<<"学生1姓名:"<< student1.name<<endl;
    cout<<"科目1:"<< student1.score1<<endl;
    cout<<"科目2:"<< student1.score2<<endl;
    cout<<"科目3:"<< student1.score3<<endl;
    cout<<"科目均分:"<< a<<endl;

    a= (student2.score1+student2.score2+student2.score3)/3;

    cout<<"学生2姓名:"<< student2.name<<endl;
    cout<<"科目1:"<< student2.score1<<endl;
    cout<<"科目2:"<< student2.score2<<endl;
    cout<<"科目3:"<< student2.score3<<endl;
    cout<<"科目均分:"<< a<<endl;
    return 0;
    }

    -----------------------------------------------------------------------------------------------

    C++完成对.txt文件读取,打印所有内容到屏幕上:

    #include <iostream>
    #include <fstream>
    #include <string>
    using namespace std;
    int main()
    {
    ifstream in;
    in.open("d:\haha.txt");//文件每层间一定要用双斜杠隔开
    if(!in)
    {
    cout<<"出错啦"<<endl;
    return 1;
    }
    char ch;
    while(!in.eof())
    {
    in.read(&ch,1);
    get(a)


    cout<<ch;}
    }
    in.close();
    return 0;
    }

    -----------------------------------------------------------------------------------------------

    读出.txt文件中学生姓名,分数,平均分

    txt文件只需要带有空格,该程序即可区分,txt文件内容如下{

    小羊 65 69 68
    小李 94 89 98

    }

    #include <iostream>
    #include <fstream>

    using namespace std;
    struct Student
    {

    char name[20];
    int s1;
    int s2;
    int s3;
    int average;
    };
    const int N=2;
    int main( )
    {

    int i, stuNum=0;
    Student stu[N];
    ifstream infile("haha.txt",ios::in);
    if(!infile)
    {
    cout<<"open error!"<<endl;
    return 0;
    }
    i=0;
    while(!infile.eof())
    {
    infile>>stu[i].name>>stu[i].s1>>stu[i].s2>>stu[i].s3;
    stu[i].average=(stu[i].s1+stu[i].s2+stu[i].s3)/3;
    ++stuNum;
    ++i;
    }
    infile.close();
    //display
    for(i=0; i<stuNum; ++i)
    {
    cout<<stu[i].name<<endl;
    cout<<"三科分别为:"<<stu[i].s1<<" "<<stu[i].s2<<" "<<stu[i].s3<<" "<<endl;
    cout<<"平均分:"<<stu[i].average<<endl;

    }
    return 0;
    }

  • 相关阅读:
    团队沟通利器之UML——活动图
    Ninject对Web Api的支持问题
    关于分布式系统的数据一致性问题
    ASP.NET Web开发框架 查询
    用泛型的IEqualityComparer<T> 去除去重复项
    数据库连接监控组件,避免日常开发中因为数据库连接长时间占用或业务完成后忘记关闭连接所带来的数据库问题
    认识项目经理
    状态模式(State Pattern)
    Django框架学习通用视图
    MS CRM 2011 Schedule Service Activities
  • 原文地址:https://www.cnblogs.com/nyc1893/p/4562860.html
Copyright © 2020-2023  润新知