C++ stage 01 notes 08 [structure (basic concepts, definition and use, array, pointer, nesting, const use)]

C++| Ingenious work from 0 to 1 introductory programming [video + courseware + notes + source code]

table of Contents

8 structure

8.1 Basic concepts of structure

8.2 Structure definition and use

Example

8.3 Structure array

Example

8.4 Structure pointer

Example

8.5 Structure Nested Structure

Example

8.6 Structure as function parameter

Example

8.7 Scenarios for using const in the structure

Example

8.8 Structure case

8.8.1 Case 1

8.8.2 Case 2

---

8 structure

8.1 Basic concepts of structure

The structure is a user -defined data type , allowing users to store different data types.

C++ built-in data types: int, float, double, string, char...

8.2 Structure definition and use

grammar:struct 结构体名 { 结构体成员列表 }；

There are 3 ways to create variables through structures :

1. struct structure name variable name

2. struct structure name variable name = {member 1 value, member 2 value...}

3. When defining the structure, create variables by the way

Example

  

#include <iostream>
#include <string>
using namespace std;

//1、创建学生数据类型：学生包括(姓名、年龄、分数)
//自定义数据类型，一些数据类型集合组成的一个类型(其实就是内置数据类型的集合)。
//语法：struct 类型名称 { 成员列表 }

//结构体定义
struct student //结构体名称-见名知意
{
	//成员列表---学生属性
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
} stu3;	//结构体变量创建方式3

//2、通过学生类型创建具体学生（共有3种方式）

//2.1、struct student s1; // s1就是变量名
//2.2、struct student s2 = {...}; // 创建变量的同时赋初值
//2.3、在定义结构体时顺便创建结构体变量

int main()
{
	//【结构体变量创建方式1】2.1、struct student s1;
	struct student stu1; //创建结构体时，struct关键字可以省略
	//给stu1属性赋值，通过“.”访问结构体变量中的属性
	stu1.name = "张三";
	stu1.age = 18;
	stu1.score = 100;
	//方式1打印输出
	cout << "姓名：" << stu1.name << " 年龄：" << stu1.age << " 分数：" << stu1.score << endl;

	//【结构体变量创建方式2】2.2、struct student s2 = {...};
	struct student stu2 = {"李四", 19, 60};
	//方式2打印输出
	cout << "姓名：" << stu2.name << " 年龄：" << stu2.age << " 分数：" << stu2.score << endl;

	//【结构体变量创建方式3】在定义结构体时顺便创建结构体变量，不建议使用
	stu3.name = "王五";
	stu3.age = 18;
	stu3.score = 80;
	//方式3打印输出
	cout << "姓名：" << stu3.name << " 年龄：" << stu3.age << " 分数：" << stu3.score << endl;

	system("pause");
	return 0;
}

• Summary 1: The keyword when defining a structure is struct, which cannot be omitted.
• Summary 2: When creating structure variables, the keyword struct can be omitted.
• Summary 3: Structure variables use the operator ``.'' to access members.

8.3 Structure array

Function: Put the custom structure into the array to facilitate maintenance .

grammar:struct 结构体名 数组名[元素个数] = { {} , {} , ... {} }

Example

 

#include <iostream>
#include <string>
using namespace std;

//结构体数组

//1、定义结构体
struct student
{
	//成员列表
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
};

int main()
{
	//2、创建结构体数组
	struct student stuArray[3] =
	{
		{"张三", 18, 88},
		{"李四", 28, 99},
		{"王五", 38, 66}
	};

	//3、给结构体数组中的元素赋值
	stuArray[2].name = "赵六";
	stuArray[2].age = 66;
	stuArray[2].score = 99;

	//4、遍历结构体数组
	for (int i = 0; i < 3; i++)
	{
		cout << "姓名：" << stuArray[i].name
			 << "\t年龄：" << stuArray[i].age
			 << "\t分数：" << stuArray[i].score << endl;
	}

	system("pause");
	return 0;
}

8.4 Structure pointer

Role: Access the members of the structure through the pointer . // Use structure pointers to manipulate structure members .

• Using the operator-> can structure pointer access structure properties .

Example

 &s: Take the address of s and return the pointer of the data type corresponding to s.

 

#include <iostream>
#include <string>
using namespace std;

//结构体指针

//1、定义学生结构体
struct student
{ //成员列表
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
};

int main()
{
	//2、创建学生结构体变量
	// struct student stu = {"张三", 25, 100}; //struct可以省略
	student stu = {"张三", 25, 99}; //struct可以省略

	//3、通过指针指向结构体变量
	// struct student *p = &stu; //struct可以省略
	student *p = &stu; //struct可以省略

	//结构体变量，可以直接通过.来访问
	cout << "姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;

	//4、通过指针访问结构体变量中的数据
	p->score = 66; //指针通过 -> 操作符可以访问成员

	//通过结构体指针访问结构体中的属性，需要利用“->”
	cout << "姓名：" << p->name << "\t年龄：" << p->age << "\t分数：" << p->score << endl;

	system("pause");
	return 0;
}

• Summary: The structure pointer can access the members of the structure through the -> operator.

8.5 Structure Nested Structure

Role: A member of a structure can be another structure .

For example: each teacher tutors a student, and a teacher’s structure records a student’s structure.

Example

#include <iostream>
#include <string>
using namespace std;

// 结构体嵌套结构体

//定义学生结构体
struct student
{
	//成员列表
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
};

//定义教师结构体
struct teacher
{
	//成员列表
	int id;				//教师编号
	string name;		//教师姓名
	int age;			//教师年龄
	struct student stu; //教师辅导的学生 子结构体
};

int main()
{
	//创建教师结构体变量
	struct teacher t1;
	t1.id = 333;
	t1.name = "老王";
	t1.age = 50;

	t1.stu.name = "张三";
	t1.stu.age = 18;
	t1.stu.score = 99;

	cout << "教师编号：" << t1.id
		 << "\t教师姓名：" << t1.name
		 << "\t教师年龄：" << t1.age
		 << "\t教师辅导的学生姓名：" << t1.stu.name
		 << endl;

	cout << "学生姓名：" << t1.stu.name
		 << "\t学生年龄：" << t1.stu.age
		 << "\t学生考试分数：" << t1.stu.score
		 << endl;

	system("pause");
	return 0;
}

Summary: In a structure, you can define another structure as a member to solve practical problems.

8.6 Structure as function parameter

Role: Pass the structure as a parameter to the function.

There are two ways of delivery:

1. Pass by value

2. Address pass

Example

#include <iostream>
#include <string>
using namespace std;

//定义学生结构体
struct student
{
	//成员列表
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
};

//打印学生信息函数
//1、值传递
void printStudent1(struct student stu) //(student stu)
{
	stu.age = 28; //修改属性
	cout << "1子函数中\t姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;
}

//2、地址传递
void printStudent2(struct student *stu) //用指针*stu接收地址
{
	stu->age = 28;
	cout << "2子函数中\t姓名：" << stu->name << "\t年龄：" << stu->age << "\t分数：" << stu->score << endl;
}

int main()
{
	//结构体作函数参数
	//将学生传入到一个参数中，打印学生身上的所有信息

	//创建结构体变量
	student stu = {"张三", 18, 100};
	stu.name = "李四";
	stu.age = 20;
	stu.score = 85;

	//1、值传递
	printStudent1(stu);
	cout << "1主函数中\t姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;

	cout << endl;

	//2、地址传递
	printStudent2(&stu);
	cout << "2主函数中\t姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;

	system("pause");
	return 0;
}

Summary: If you don't want to modify the data in the main function, pass it by value, otherwise pass it by address.

8.7 Scenarios for using const in the structure

Role: Use const to prevent misoperation .

Example

#include <iostream>
#include <string>
using namespace std;

//学生结构体定义
struct student
{
	//成员列表
	string name; //姓名
	int age;	 //年龄
	int score;	 //分数
};

/*
 * 值传递
 * stu.age = 150;//mian中的stu变量不会改变
 * 会将实参中的数据拷贝一份，放在形参s上
 * 无论形参如何改变，都不会影响实参
 * 1、实参有很多属性，每个属性都拷贝一份，拷贝的数据量很大
 * 2、假设一个学校有成千上万个人，每个人都调用printStudent函数，拷贝成千上万份数据，
 *    这样数据量就会非常大，占用内存空间大
 */
void printStudent1(student stu) // (struct student stu) 省略struct
{
	stu.age = 150; //mian中的stu变量不会改变
	cout << "姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;
}

// void printStudent3(const student stu) // (struct student stu) 省略struct
// {
// 	stu.age = 150; //mian中的stu变量不会改变
// 	cout << "姓名：" << stu.name << "\t年龄：" << stu.age << "\t分数：" << stu.score << endl;
// }

//将函数中的形参改为指针，可以减少内存空间的使用，而且不会复制新的副本出来（值传递会复制整个结构体元素）
//const使用场景
void printStudent2(const student *stu) //加const防止函数体中的误操作 指针*stu节省空间，一个指针占4个字节内存
{
	//stu->age = 100; //操作失败,因为加了const修饰,常量指针无法修改指针指向的值,只能读不能写。防止age的值会被修改
	//加入const之后，一旦有修改的操作就会报错，可以防止我们的误操作（加“const”防止误操作）
	cout << "姓名：" << stu->name << "\t年龄：" << stu->age << "\t分数：" << stu->score << endl;
}

int main()
{
	//创建结构体变量
	student stu = {"张三", 18, 100};

	printStudent1(stu); //值传递

	//通过函数打印结构体变量信息
	printStudent2(&stu); //传入地址，在函数中指针接收地址

	cout << "main()中 张三的年龄为：" << stu.age << endl;

	system("pause");
	return 0;
}

8.8 Structure case

8.8.1 Case 1

Case description:

The school is doing a final project. Each teacher leads 5 students, and there are 3 teachers in total. The requirements are as follows:

• Design the structure of students and teachers. In the structure of the teacher, there are the teacher's name and an array of 5 students as members;
• Student members have names and test scores, create an array to store 3 teachers, and assign values ​​to each teacher and the students they bring through functions;
• Finally print out the teacher data and the student data brought by the teacher.

#include <iostream>
#include <string>
#include <ctime>
using namespace std;

//学生结构体
struct Student
{
    string sName; //姓名
    int score;    //分数
};

//老师结构体
struct Teacher
{
    string tName;             //姓名
    struct Student sArray[5]; //学生数组
};

//创建数据——给老师和学生赋值的函数
void allocateSpace(struct Teacher tArray[], int len) //struct可以省略
{
    string tName = "老师"; // Teacher_
    string sName = "学生";
    string nameSeed = "ABCDE";
    //给老师进行赋值
    for (int i = 0; i < len; i++)
    {
        tArray[i].tName = tName + nameSeed[i]; //老师姓名
        //通过循环给每名老师所带的学生进行赋值
        for (int j = 0; j < 5; j++)
        {
            tArray[i].sArray[j].sName = sName + nameSeed[j];
            tArray[i].sArray[j].score = rand() % 61 + 40; // 0~60 40~100
            // rand() % 60：0~59
        }
    }
}

//打印数据——打印所有信息
void printTeachers(struct Teacher tArray[], int len) //struct可以省略
{
    for (int i = 0; i < len; i++)
    {
        cout << "老师的姓名：" << tArray[i].tName << endl;
        for (int j = 0; j < 5; j++)
        {
            cout << "\t学生姓名：" << tArray[i].sArray[j].sName << "；考试分数：" << tArray[i].sArray[j].score << endl;
        }
    }
}

int main()
{
    //利用系统时间，产生随机数
    srand((unsigned int)time(NULL)); //随机数种子 头文件 #include <ctime>

    //1、创建3名老师的数组
    struct Teacher tArray[3]; //老师数组 struct可以省略

    //2、通过函数给3名老师的信息赋值，并给老师带的学生信息赋值
    cout << "string类型所占内存空间为：" << sizeof(string) << endl; //string类型所占内存空间为：32
    cout << "int类型所占内存空间为：" << sizeof(int) << endl; //int类型所占内存空间为：4
    cout << "sizeof(tArray)：" << sizeof(tArray) << endl;            //sizeof(tArray)：696
    cout << "sizeof(Teacher)：" << sizeof(Teacher) << endl;          //sizeof(Teacher)：232
    cout << "sizeof(Student)：" << sizeof(Student) << endl;          //sizeof(Student)：40
    cout << "sizeof(tArray[0])：" << sizeof(tArray[0]) << endl;      //sizeof(tArray[0])：232
    int len1 = sizeof(tArray) / sizeof(Teacher);                     //计算数组长度
    cout << "len1：" << len1 << endl;                                //len1：3
    int len2 = sizeof(tArray) / sizeof(tArray[0]);                   //计算数组长度
    cout << "len2：" << len2 << endl;                                //len2：3
    allocateSpace(tArray, len2);                                     //创建数据

    //3、打印所有老师及所带的学生信息
    printTeachers(tArray, len2); //打印数据

    system("pause");
    return 0;
}

8.8.2 Case 2

Case description:

Design a hero structure, including member names, ages, and genders; create a structure array, and store 5 heroes in the array.

Through the bubble sort algorithm , the heroes in the array are sorted in ascending order by age , and the sorted result is finally printed.

The information of the five heroes is as follows:

{"Liu Bei", 23, "Male"},
{"Guan Yu", 22, "Male"},
{"Zhang Fei", 20, "Male"},
{"Zhao Yun", 21, "Male"},
{ "Diao Chan", 19, "Female"},

#include <iostream>
#include <string>
using namespace std;

//1、设计英雄结构体
//英雄结构体
struct hero
{
    string name; //姓名
    int age;     //年龄
    string sex;  //性别
};

//冒泡排序——实现年龄升序排列
void bubbleSort(struct hero arr[], int len) //struct可省
{
    for (int i = 0; i < len - 1; i++)
    {
        for (int j = 0; j < len - 1 - i; j++)
        {
            //如果j下标的元素年龄大于j+1下标的元素的年龄，交换两个元素
            if (arr[j].age > arr[j + 1].age)
            {
                struct hero temp = arr[j]; //临时数据
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }
}

//打印数组——打印排序后数组中的信息
void printHeros(struct hero arr[], int len) //struct可省
{
    for (int i = 0; i < len; i++)
    {
        cout << "姓名：" << arr[i].name << "\t性别：" << arr[i].sex << "\t年龄：" << arr[i].age << endl;
    }
}

//1、设计英雄结构体
//2、创建数组存放5名英雄
//3、对数组进行排序，按照年龄进行升序排序
//4、将排序后的结果打印输出
int main()
{
    //2、创建数组存放5名英雄
    struct hero heroArray[5] =
        {
            {"刘备", 23, "男"},
            {"关羽", 22, "男"},
            {"张飞", 20, "男"},
            {"赵云", 21, "男"},
            {"貂蝉", 19, "女"}};

    cout << "string类型所占内存空间为：" << sizeof(string) << endl;   //string类型所占内存空间为：32
    cout << "int类型所占内存空间为：" << sizeof(int) << endl;         //int类型所占内存空间为：4
    cout << "sizeof(heroArray)：" << sizeof(heroArray) << endl;       //sizeof(heroArray)：360
    cout << "sizeof(heroArray[0])：" << sizeof(heroArray[0]) << endl; //sizeof(heroArray[0])：72
    cout << "sizeof(hero)：" << sizeof(hero) << endl;                 //sizeof(hero)：72

    int len1 = sizeof(heroArray) / sizeof(heroArray[0]); //整体所占空间大小/单个元素所占空间大小
    cout << "len1：" << len1 << endl;                    //len1：5
    int len2 = sizeof(heroArray) / sizeof(hero);         //获取数组元素个数
    cout << "len2：" << len2 << endl;                    //len1：5

    cout << "排序前，数组打印：" << endl;
    for (int i = 0; i < len1; i++)
    {
        cout << "姓名：" << heroArray[i].name << "\t性别：" << heroArray[i].sex << "\t年龄：" << heroArray[i].age << endl;
    }

    //3、对数组进行排序，按照年龄进行升序排序
    bubbleSort(heroArray, len1); //排序

    cout << "排序后，数组打印：" << endl;

    //4、将排序后的结果打印输出
    printHeros(heroArray, len1); //打印

    system("pause");
    return 0;
}

3 points of this case exercise:

1. How to write structure array
2. Bubble sort // Use "bubble sort" to sort the structure array
3. The writing method of structure incoming function, the writing method of operation structure