超載<<在C基於模板對象++

Question

類

我試圖超載<<爲包含對象的數組MemSet模板...但如果我試圖做的MemSetMemSet則超載不起作用

///// SetMain.cpp 
int main() 
{ 
    MemSet<Record> mset(10),mset2(10); 
    MemSet< MemSet<Record> >mm1(10); 
    Record r,r2,r3,r4,r5,r6; 


cin>>r; 
cin>>r2; 
cin>>r3; 
mset.add(r); 
mset.add(r2); 
mset.add(r3); 
cout<<" Memset 2:"<<endl; 

cin>>r4; 
cin>>r5; 
cin>>r6; 
mset2.add(r); 
mset2.add(r2); 
mset2.add(r3); 


if(mset == mset2) 
cout<<"Both memset are equal"<<endl; 
else 
cout<<"Memsets are not equal"<<endl; 

mm1<<mset; 
//mset2>>mm2; 

if(mm1 == mm2) 
cout<<"Both memset of memset are equal"<<endl; 
else 
cout<<"Memset of memset are not equal"<<endl; 


return 0; 
} 

這裏是MEMSET類....

////Set and MemSet class 

     template <class T> 
    class Set 
    { 
      public : 
       virtual bool find(T x) =0; 
       virtual bool add(T x) =0; 
    }; 


template<class T> 
class MemSet : public Set<T> 
{ 
    public: 
    T *array; 
    int arraysize,currentsize; 
template <class T1> 
friend istream& operator >> (istream& s,MemSet<T1>& m); 
template <class T1> 
friend ostream& operator << (ostream& s,MemSet<T1>& m);   

}; 


template <class T> 
ostream& operator << (ostream& s,MemSet<T>& m) 
{ 
    for(int i=0;i<m.getCurrentsize();i++) 
    s << m.array[i]; 


    return s; 
    } 

Record類

////Record class 
class Record 
     { 
      public: 
      int age; 
      std::string first_name,last_name; 


friend istream& operator >> (istream& s,Record& r); 
friend ostream& operator << (ostream& s,Record& r);   
}; 

istream& operator >> (istream& s,Record& r) 
        {  
         s>>r.age; 
         s>>r.first_name; 
         s>>r.last_name; 
         return s; 
         }    
ostream& operator << (ostream& out,Record& r) 
        { 
         out << r.age ; 
         out << r.first_name ; 
         out << r.last_name ; 
         return out ; 
         } 

這些都是使用語法我得到

SetMain.cpp: In function ‘int main()’: 
SetMain.cpp:36:6: error: no match for ‘operator<<’ in ‘mm1 << mset’ 
SetMain.cpp:36:6: note: candidates are: 
Set.cpp:111:10: note: template<class T> std::ostream& operator<<(std::istream&, MemSet<T>&) 
Record.cpp:46:10: note: std::ostream& operator<<(std::ostream&, Record&) 
Record.cpp:46:10: note: no known conversion for argument 1 from ‘MemSet<MemSet<Record> >’ to ‘std::ostream& {aka std::basic_ostream<char>&}’ 

Answer 1

錯誤添加 'MSET' 到 'MM1' mm1 << mset代替mm1.add(mset)可以通過做：

template<typename T> 
MemSet<T>& operator << (MemSet<T>& m, const T& v) { 
    m.add (v); 
    return m; 
} 

或者相當於

template<typename T> 
class MemSet: public Set { 
public: 
    MemSet& operator << (const T& v) { 
     add (v); 
     return *this; 
    } 
}; 

但是，在添加「句法su」時，我會小心CuteProgramming噶「這樣的代碼。

Answer 2

下面是一個例子，它提供了類似於我想要的行爲。您需要擴展這些類以提供比較。

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

class Record 
{ 
    friend istream& operator>> (istream& s, Record& r); 
    friend ostream& operator<< (ostream& s, Record& r); 

public: 
    int GetAge(void) const { return m_nAge; } 
    const std::string& GetFirstName(void) const { return m_strFirstName; } 
    const std::string& GetLastName(void) const { return m_strLastName; } 

private: 
    int m_nAge; 
    string m_strFirstName; 
    string m_strLastName; 
}; 

istream& operator>> (istream& s, Record& r) 
{ 
    s >> r.m_nAge; 
    s >> r.m_strFirstName; 
    s >> r.m_strLastName; 
    return s; 
} 

ostream& operator<< (ostream& s, Record& r) 
{ 
    s << "Individual: " << r.m_strLastName << ", " 
     << r.m_strFirstName << " (" << r.m_nAge << " years)" << endl; 
    return s; 
} 

template<class T> 
class MemSet : public vector<T> 
{ 
    template<class T1> 
    friend ostream& operator<< (ostream& s, MemSet<T1>& m); 

    template<class T1> 
    friend MemSet<T1>& operator<< (MemSet<T1>& m, T1& t); 
}; 

template<class T> 
ostream& operator<< (ostream& s, MemSet<T>& m) 
{ 
    for (typename vector<T>::iterator it = m.begin(); it != m.end(); ++it) 
     s << *it; 
    return s; 
} 

template<class T> 
MemSet<T>& operator<< (MemSet<T>& m, T& t) 
{ 
    m.push_back(t); 
    return m; 
} 

int main(int argc, char** argv) 
{ 
    MemSet<Record> mset1, mset2; 
    MemSet< MemSet<Record> > mm; 
    Record temp; 

    cin >> temp; mset1.push_back(temp); 
    cin >> temp; mset1.push_back(temp); 
    cin >> temp; mset1.push_back(temp); 

    cin >> temp; mset2.push_back(temp); 
    cin >> temp; mset2.push_back(temp); 
    cin >> temp; mset2.push_back(temp); 

    cout << mset1 << endl; 

    mm << mset1; 
    mm << mset2; 

    cout << mm << endl; 

    return 0; 
}