[c++]Problem z razredi

	
int main(){
PEmployee one;
	PEmployee two;

	PEmployee three("James", 1000);
	three.set_salary(1200);
	
	cout << three.get_name() << "  " << three.get_salary() << '\n';

	cout << "Press any key to countinue...";
	cin.get();
return 0;}

#ifndef Person_H
#define Person_H

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


class Person{

public:
	Person();
	Person(string pname, int page);
	string get_name() const;
	void get_age() const;
private:
	string name;
	int age; // 0 if unknown

};

#endif 

#ifndef PEmployee_H
#define PEmployee_H

#include "Person.h"
#include <iostream>
#include <string>
using std::string;

class PEmployee
{
public:
	PEmployee();
	PEmployee(string employee_name, double initial_salary);
	void set_salary(double new_salary);
	double get_salary() const;
	string get_name() const;
private:
	Person person_data;
	double salary;
};

#endif

#include "Person.h"

using namespace std;


Person::Person(){

	name = "noname";
	age = 0;

}

Person::Person(string pname, int page){

	name = pname;
	age = page;

}

string Person::get_name() const{

	return name;

}

void Person::get_age() const{

	cout << age << '\n';

}

#include "PEmployee.h"

using namespace std;

PEmployee::PEmployee(){
	
	Person person_data();
	salary = 0;
	
}

PEmployee::PEmployee(string employee_name, double initial_salary){

	Person person_data(employee_name, 0);
	salary = initial_salary;

}



void PEmployee::set_salary(double new_salary){
	
	salary = new_salary;
}


double PEmployee::get_salary() const{
	
	return salary;

}

string PEmployee::get_name() const{

	return person_data.get_name();  
}

noname 1200

class Employer : public Person
{
   // poklices se base class kontruktor
   public Employer(string name, int age, int es) : Person(name,age) 
   {
       salary = es;
   }
}

Employer e1("vesoljc", 33, 5000);  // notice the funny part ;)
cout << e1.getname() << e1.getsalary();

killa bee je 10. nov 2014 ob 17:02 izjavil:

to se mi pa res ne zdi logično. String in int spremenljivke se ko se konstruktor izvede ostanejo za stalno. Če je pa tipa Person pa ne ostane za stalno.
SEpravi se ko se konstruktor

PEmployee three("James", 1000);

izvede se v bistvu izvedeta 2 konstruktorja. Person() ter Person(James,0)?

Person person_data(employee_name, 0);

killa bee je 11. nov 2014 ob 11:45 izjavil:

ja nastavi, ampak se enkrat, ta instanca je zacasna in se unici ko se employer konstruktor konca

Zakaj bi se pa uničila?
Kdo pa potem nastavi spremenljivko -name na "noname"?

smoke je 11. nov 2014 ob 11:56 izjavil:

killa bee je 10. nov 2014 ob 14:28 izjavil:

PEmployee.cpp

#include "PEmployee.h"
 
 using namespace std;
 
 PEmployee::PEmployee(){
 
 Person person_data();
 salary = 0;
 
 }
 

Person person_data(); NI lokalna spremenljivka, temveč deklaracija funkcije z imenom person_data, ki ne sprejme nobenega parametra in vrača objekt Person.

mn je 11. nov 2014 ob 12:22 izjavil:

@smoke Sem preveril in drži. Zanimivo, tega res nisem vedel (ali pa že pozabil).

This is known as "C++'s most vexing parse". Basically, anything that can be interpreted by compiler as a declaration will be interpreted as a declaration.

Ko pa pokličeš get_name() pa dostopaš do razredne spremenljivke, ki je bila nastavljena ko se je poklical privzeti konstruktor RAZREDNE spremenljivke person_data.

Exercise P5.1. Implement all member functions of the following class:
class Person
{
public:
Person();
Person(string pname, int page);
void get_name() const;
void get_age() const;
private:
string name;
int age; // 0 if unknown
};
------------------------------------------------------------------------------------------
Exercise P5.2. Implement a class PEmployee that is just like the Employee class except
that it stores an object of type Person as developed in Exercise P5.1.
class PEmployee
{
public:
PEmployee();
PEmployee(string employee_name, double initial_salary);
void set_salary(double new_salary);
double get_salary() const;
string get_name() const;
private:
Person person_data;

ampak se bo poklical konstruktor, ki sprejema en parameter tipa int.

class A {
 public:
  A(int x) : x_(x) {}
 
 private:
  int x_;
};

Senitel je 12. nov 2014 ob 10:19 izjavil:

Lahko ne napišeš privzetega konstruktorja, ampak ga bo compiler še vedno avtomatsko generiral. Lahko ga pa prepoveš v stilu:

 class A {
 private:
 A() {}
 
 int x_;
 public:
 A(int x) : x_(x) {}
 };
 

Ampak to nima direktne veze s tvojim problemom, da ne razumeš kako sploh deluje scope in kako se instancirajo objekti.

class A {
 public:
  A(const A&) {}
};

std::cout << "Ime Funkcije" << std::endl;