[英]how can i use aggregation while separating my classes into a .h and .cpp files?
我的教授給了我一個任務,讓我使用 2 個類之間的聚合來制作一個程序,同時將這些類分成.h
和.cpp
文件。
包含 class 聲明的 header 文件:
#include <iostream>
#include <string>
using namespace std;
class medicalCompany {
private:
string ceoName;
string email;
string phoneNumber;
string locate;
public:
medicalCompany();
void Name(string n);
void mail(string m);
void phone(string p);
void location(string l);
~medicalCompany();
};
class origin {
private:
medicalCompany country;
public:
origin();
void address();
~origin();
};
和我的.cpp 文件:
#include <iostream>
#include "function.h"
#include <string>
using namespace std;
medicalCompany::medicalCompany() {
cout << "OUR COMPANY IS GLAD TO BE OF SERVICE !" << endl;
cout << "****************************************************" << endl;
}
void medicalCompany::Name(string n){
ceoName = n;
cout << "OUR CEO IS " << endl;
cout<< ceoName << endl;
cout << "****************************************************" << endl;
}
void medicalCompany::mail(string m) {
email = m;
cout << "USE OUR EMAIL TO CONTACT US : " << endl;
cout<< email << endl;
cout << "****************************************************" << endl;
}
void medicalCompany::phone(string p) {
phoneNumber = p;
cout << "THIS IS OUR CUSTOMER SERVICE PHONE NUMBER " << endl;
cout<< phoneNumber << endl;
cout << "****************************************************" << endl;
}
void medicalCompany::location(string l) {
locate = l;
cout << " OUR COMPANY IS LOCATED IN " << endl;
cout << locate << endl;
cout << "****************************************************" << endl;
}
medicalCompany::~medicalCompany() {
cout << "thanks for trusting our company ^_^" << endl;
cout << "****************************************************" << endl;
}
origin::origin() {
cout<< "constructor 2"<<endl;
}
void origin::address() {
cout << country.location;
}
origin::~origin() {
cout << "bye" << endl;
}
這兩個類在我的main.cpp
文件中使用:
#include <iostream>
#include <string>
#include "function.h"
using namespace std;
int main() {
medicalCompany o;
o.Name("jack");
o.mail("ouremail@company.com");
o.phone("2342352134");
o.location("Germany");
origin o2;
return 0;
}
我遇到了這個錯誤:
Severity Code Description Project File Line Suppression State
Error C3867 'medicalCompany::location': non-standard syntax; use '&' to create a pointer to member CP2_HW c:\function.cpp 41
您可以:
將void origin::address(){cout << country.location;}
替換為 void void origin::address(){country.location();}
或者by void origin::address(){cout << country.locate;}
如果您將locate
成員作為公共變量。
另外,幾點說明:
通常您希望避免暴露私有成員,因此應該首選第一種解決方案。
using namespace std;
通常被認為是不好的做法,應該避免,因為可能的風險成本不會超過不必鍵入std::cout
的好處(有關更多信息,請參見此問題)
在命名約定方面,我會交換locate
和location
的名稱: location 應該是成員變量和locate
獲取位置的動作(函數)。
更喜歡使用構造函數和初始化列表而不是 getter/setter。
您的 output 格式應該與您的類的邏輯完全分開,例如為您的 class 實現<<
運算符。
按照這個邏輯,您的代碼應該更像:
#include <iostream>
#include <string>
class medicalCompany {
private:
std::string _ceoName;
std::string _email;
std::string _phoneNumber;
std::string _location;
public:
// Constructor
medicalCompany(std::string name, std::string email, std::string phone, std::string location):
_ceoName(name),
_email(email),
_phoneNumber(phone),
_location(location)
{}
friend ostream& operator<<(ostream& os, const medicalCompany& dt);
};
對於 ostream 運營商:
ostream& operator<<(ostream& os, const medicalCompany& co)
{
os << co._ceoName << " " co._phoneNumber << ' ' << co._email;
return os;
}
這將允許在您的 main 中編寫這樣的代碼:
int main() {
medicalCompany o("jack", "ouremail@company.com","2342352134","Germany")
std::cout << o << std::endl;
return 0;
}
該代碼不起作用,您必須對其進行編輯以滿足您的格式要求,但您有想法:)祝您好運!
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