我有一个类层次结构,如下面的示例所示,其中State
包含ZipCode
的列表和City
的列表,每个列表均包含指向ZipCode
的指针。
目标是无需更新ZipCode
即可更新City
(或创建City
的新实例)。
下面的C++代码满足了这一要求,但是它使用了指针,由于this和that,我更希望避免使用指针。 如何重新设计此[naive]实现,使其不依赖于指针? 谢谢您的帮助!
编辑:更新了以下代码,以使用boost::shared_ptr
代替原始指针。请注意,State
,City
和ZipCode
只是示例名称,它们被证明是较差的选择名称(我可以选择“A”,“B”和“C”),因为实际代码允许City
共享ZipCode
。
#include <iostream>
#include <vector>
#include <boost/shared_ptr.hpp>
using namespace std;
/**
* Zone Improvement Plan (ZIP) code
*/
class ZipCode {
public:
ZipCode() : code_(0), plus4_(0) {}
ZipCode(int code, int plus4 = 0) : code_(code), plus4_(plus4) {}
virtual ~ZipCode() {};
int code() const { return code_; }
int plus4() const { return plus4_; }
void set_code(int code) { code_ = code; }
void set_plus4(int plus4) { plus4_ = plus4; }
private:
int code_;
int plus4_;
};
typedef boost::shared_ptr<ZipCode> ZipPtr;
/**
* City points to one or more zip codes
*/
class City {
public:
const vector<ZipPtr>& zip() const { return zip_; }
void add_zip_ptr(const ZipPtr x) { if (x != NULL) zip_.push_back(x); }
private:
// TODO: this vector should be a hash set
vector<ZipPtr> zip_;
};
/**
* State contains cities, each of which has pointers to
* zip codes within the state.
*/
class State {
public:
const vector<City>& city() const { return city_; }
const vector<ZipPtr>& zip() const { return zip_; }
const ZipPtr zip_of(int code) const {
for (size_t i = 0; i < zip_.size(); i++) {
if (zip_[i]->code() == code) {
return zip_[i];
}
}
return ZipPtr();
}
void add_city(const City& x) { city_.push_back(x); }
void add_zip(int code) { zip_.push_back(ZipPtr(new ZipCode(code))); }
private:
// TODO: these vectors should be hash sets
vector<City> city_;
vector<ZipPtr> zip_;
};
int main() {
State texas;
City dallas, houston;
// create state ZIPs
texas.add_zip(75380);
texas.add_zip(75381);
texas.add_zip(77219);
texas.add_zip(77220);
// point city ZIPs to the ones we just created
dallas.add_zip_ptr(texas.zip_of(75380));
dallas.add_zip_ptr(texas.zip_of(75381));
houston.add_zip_ptr(texas.zip_of(77219));
houston.add_zip_ptr(texas.zip_of(77220));
// print all ZIPs
cout << "ZIPs in Texas: " << endl;
const vector<ZipPtr>& zips = texas.zip();
for (size_t i = 0; i < zips.size(); i++) {
cout << " " << zips[i]->code() << endl;
}
cout << "ZIPs in Dallas, Texas: " << endl;
const vector<ZipPtr> zip_ptrs1 = dallas.zip();
for (size_t i = 0; i < zip_ptrs1.size(); i++) {
cout << " " << zip_ptrs1[i]->code() << endl;
}
cout << "ZIPs in Houston, Texas: " << endl;
const vector<ZipPtr> zip_ptrs2 = houston.zip();
for (size_t i = 0; i < zip_ptrs2.size(); i++) {
cout << " " << zip_ptrs2[i]->code() << endl;
}
// change a state ZIP...
cout << "Changing Houston's ZIP 77220..." << endl;
ZipPtr z = texas.zip_of(77220);
if (z != NULL) z->set_code(88888);
// ...and show the ZIPs of the affected city
cout << "ZIPs in Houston, Texas: " << endl;
const vector<ZipPtr> zip_ptrs3 = houston.zip();
for (size_t i = 0; i < zip_ptrs3.size(); i++) {
cout << " " << zip_ptrs3[i]->code() << endl;
}
return 0;
}
最佳答案
我认为情况是两个1:n关系
== 1:n
基于此,我认为包含的
State
vector<ZipCode> zip_;
没有声音。
我可能会做
class State {
vector< City > cities_in_state_;
};
class City {
vector< Zipcode > zips_in_city_;
};
这不需要指针。