how to use strcopy in c++

in c++ , i have a file named "Student.h"

class LinkedList {
private: 

class Student {
public:

    int stId;
    char stName [20];
    char stMajor[20];
    double stAverage;
    Student * next;

    Student() {
        next = 0;
    }

    Student(int stId, char stName [20], char stMajor[20], double stAverage) {
        this.stId = stId;
        strcopy(this.stName, stName); // there is error here !
    开发者_StackOverflow    strcopy(this.stMajor, stMajor);
        this.stAverage = stAverage;
    }

what should i do ?!

this is a pointer in C++, not a reference as in Java. Plus what you need is strcpy() not strcopy()

Try this

    strcpy(this->stName, stName); 
    strcpy(this->stMajor, stMajor);

P.S : In C++ it is always recommended to prefer std::string over C-style arrays

A much cleaner version of your code would be something like this

struct Student {

    int stId;
    std::string stName;
    std::string stMajor;
    double stAverage;
    Student * next;

    Student():stId(),stAverage(),next()//prefer initialization-list to assignment
    {
    }

    Student(int stId, const std::string &stName, const std::string &stMajor, double stAverage){
      this->stId = stId,
      this->stName = stName ,
      this->stMajor = stMajor,
      this->stAverage = stAverage;          
    }
};

what should i do ?!

You should:

• use std::string instead of raw arrays.

• use std::list instead of inventing your own (except for purpose of learning about linked lists).

• not indicate array sizes in formal arguments, like your char stName [20]; the formal argument type does not retain the size information, it just devolves to a pointer type.

• generally avoid using this directly.

• generally use initializer lists instead of assignments in the constructor body.

Cheers & hth.,

I think you mean the strcpy function (without o).

Can you not use std::string?

string s1, s2 = "example";
s1 = s2;

Anyway, the problem is that in C++ this returns a pointer, therefore this.stId is wrong, the correct form would be this->stId, or alternatively (*this).stId.

this is a pointer, not a reference, so you have to use pointer dereferencing operators:

    strcpy(this->stName, stName);

or

    strcpy((*this).stName, stName);

Moreover, I don't recommend using char[20] as a datatype for student names - that is very prone to buffer overflow errors. You could overcome this by using strncpy

    strcpy(this->stName, stName, 19);
    this->stName[20]=0;

But the most convenient way is using std::string, which can be conveniently copied by assignment.

And last, if you chose some convention for member variable names, you could just refer to them without this. Eg:

class Student {
public:

    std::string m_stName;

...
    Student(int stId, std::string stName, ...) {
         m_stName=stName;

or even (using initializers):

Student(int stId, std::string stName, ...) : m_stName(stName) {
     m_stName=stName;