这是一个简单的C ++项目,具有两种设计模式:单例和工厂,sigleton也是模板化类,还包括接口(IHash)和类(Hash1)。
一个简单的工厂类(HashFactory)创建一个sigleton(Hash1); Hash1继承了接口IHash,理想情况下,我具有Hash1,Hash2 .. HashN。
在编译时出现错误,这是什么问题?
g++ main.cpp
main.cpp: In static member function ‘static IHash* HashFactory::get(int)’:
main.cpp:11:15: error: ‘static T& Singleton<T>::getInstance() [with T = Hash1]’ is inaccessible
static T &getInstance() {
^
main.cpp:76:50: error: within this context
if (type == 1)return &Hash1::getInstance();
^
剪切并粘贴以下代码进行编译:
#include <iostream>
using namespace std;
///////////////////////////////////////////////
//Class Singleton
template<class T>
class Singleton {
public:
static T &getInstance() {
if (!_instanceSingleton) {
_instanceSingleton = new T();
}
return *_instanceSingleton;
}
private:
static T *_instanceSingleton;
};
template<class T> T *Singleton<T>::_instanceSingleton = 0;
/////////////////////////////////////////////////
//Interface IHash
class IHash {
public:
void function1() {
cout << "function1";
}
virtual void recordHash(bool b) = 0;
~IHash() {
dispose();
}
private:
void dispose() {
cout << "dispose\n";
}
};
///////////////////////////////////////////////////
//Class Hash1 is a singleton and inherits IHash
class Hash1 : public IHash, Singleton<Hash1> {
friend class Singleton<Hash1>;
public:
void recordHash(bool b);
private:
//private constructor, is a sigleton
Hash1();
};
Hash1::Hash1() {
cout << "create Hash1\n";
}
void Hash1::recordHash(bool b) {
cout << b << " recordHash\n";
}
////////////////////////////////////////////////////
//Factory for IHash
class HashFactory {
public:
static IHash *get(int type) {
if (type == 1)return &Hash1::getInstance();
// if (type == 2)return &Hash2::getInstance();
// if (type == 3)return &Hash3::getInstance();
return 0;
}
};
//////////////////////////////////////////////////////
int main() {
int type=1;
IHash *a = HashFactory::get(type);
a->recordHash(true);
a->function1();
return 0;
}
最佳答案
Hash1
从Singleton<Hash1>
的继承是隐式私有的。更改为
class Hash1 : public IHash, public Singleton<Hash1> {