数组赋给数组

遇到的问题：c++中一个数组传给另一个数组时不比配的问题

正确版本1：

#include "iostream"
#include"string"

using namespace std;


class Student{
private:
    char *sno,*sname;
    int age;
public:
    Student(char sno[],char sname[],int age){
        this->sno = sno;
        this->sname = sname;
        this->age = age;
        
        
    }
    void print(){
        
        cout<<"sno:"<<sno<<" sname:"<<sname<<" age:"<<age<<endl;
    }

    
    
};

int main(){
    char sno[10],sname[20];
    int age;
    while(cin>>sno>>sname>>age){
        Student A(sno,sname,age);
        A.print();
        
        return 0;
    }
    
 
}

结果

正确2:

//....
public:
    Student(char* sno,char * sname,int age){
        this->sno = sno;
        this->sname = sname;
        this->age = age;
        
        
    }
//....

数组名和指针皆为地址，不影响。但是有区别这两者，前者指的是数组的第一个元素的地址，并且是不可改的，常量指针，改了的话，那块地址的数组就无法使用，会内存溢出
；后者是可改的。

下面这个更具体：
https://segmentfault.com/q/1010000003793687
错误1:

//...
public:
    Student(char[] sno,char[] sname,int age){
        this->sno = sno;
        this->sname = sname;
        this->age = age;
//...

错误

char[] a; java风格
char a[]; c/c++风格

错误2:

class Student{
private:
    char sno[10],sname[20];
    int age;
public:
    Student(const char* sno,char * sname,int age){
        this->sno = sno;
        this->sname = sname;
        this->age = age;
        
        
    }

理由正如上面所说，数组名（地址）不可改（赋值）。
只能用 strcpy ,memcpy.

To copy the content into the array, you need to use strcpy() from string.h (char array) or memcpy() in general.

正确：

class Student{
private:
    char sno[10],sname[20];
    int age;
public:
    Student( char* sno,char * sname,int age){
        strcpy(this->sno,sno);
        strcpy(this->sname,sname);
        
//        this->sno = sno;
//        this->sname = sname;
        this->age = age;
        
        
    }

总结：
概念不清，没有细细的理解。数组和指针的熟练度太低。