#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
#include "Myvector.cpp"

class Teacher
{
public:


    Teacher(char* name, int age)
    {
        this->age = age;
        strcpy(this->name, name);
    }


    void printT()
    {
        cout << name << "------" << age << endl;
    }
private:
    int age;
    char *name;
};

int main(void)
{
    Teacher t1("t1", 31), t2(32, "t2"), t3(33, "t3");
    system("pause");
    return 0;
}

// error C2664: “Teacher::Teacher(Teacher &&)”: 无法将参数 1 从“const char [3]”转换为“char *”
// note: 从字符串文本转换将丢失 const 限定符(请参阅 /Zc:strictStrings)

今天我也遇到相同的问题
构造函数处 你输入的是一个字符串，编译器默认为const char*类型，而Teacher(char* name, int age)中声明的为 char *，类型转换出错，将char*加上const修饰符就解决啦
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
#include "Myvector.cpp"

class Teacher
{
public:


    Teacher(const char* name, int age)
    {
        this->age = age;
        strcpy(this->name, name);
    }


    void printT()
    {
        cout << name << "------" << age << endl;
    }
private:
    int age;
    const char *name;
};

int main(void)
{
    Teacher t1("t1", 31), t2(32, "t2"), t3(33, "t3");
    system("pause");
    return 0;
}

谢谢哈