爲什麼不能編譯這個C++模板代碼？

Question

這是從The C++ Programming Language第17章複製的一些示例代碼，如下所示。當我用Visual Studio 2008進行編譯時，它一直給我這個錯誤： warning C4346: 'HashMap<Key,T,H,EQ,A>::mapped_type' : dependent name is not a type 1> prefix with 'typename' to indicate a type

有沒有人有任何想法呢？ 在此先感謝！

#include <vector> 
#include <map> 
using std::vector; 
using std::pair; 
using std::iterator; 

template<class Key, class T, class H= Hash<Key>, class EQ = equal_to<Key>, class A = allocator<pair<const Key, T>>> 
class HashMap{ 
public: 
    typedef Key key_type; 
    typedef T mapped_type; 
    typedef pair<const Key, T> value_type; 
    typedef typename A::size_type size_type; 
    typedef H Hasher; 
    typedef EQ key_equal; 

    HashMap(const T& dv=T(), size_type n = 101, const H& hf = H(), const EQ& = EQ()): 
      :default_value(dv), b(n), no_of_erased(0), hash(hf), eq(e){ 
     set_load(); 
     v.reserve(max_load*b.size()); 
    } 
    template<class In> HashMap(In first, In last, 
     const T& dv=T(), size_type n=101, const H& hf=H(), const EQ& = EQ()); 

    void set_load(float m=0.7, float g=1.6){ 
     max_load = m; 
     grow = g; 
    } 
    mapped_type& operator[](const key_type& k); 

    void resize(size_type n); 
    void erase(iterator position); 
    size_type size() const { return v.size() - no_of_erased;} 
    size_type bucket_count() const { return b.size();} 
    Hasher hash_fun() const { return hash;} 
    key_equal key_eq() const { return eq;} 

private: 
    struct Entry{ 
     key_type key; 
     mapped_type val; 
     bool erased; 
     Entry* next; 
     Entry(key_type k, mapped_type v, Entry* n): 
      key(k),val(v),erased(false),next(n) {} 
    }; 

    vector<Entry> v; 
    vector<Entry*> b; 

    float max_load; 
    float grow; 
    size_type no_of_erased; 
    Hasher hash; 
    key_equal eq; 

    const T default_value; 

}; 


template<class Key, class T, class H=Hash<Key>, class EQ=equal_to<Key>, class A=allocator<pair<const Key,T>>> 
HashMap<Key,T,H,EQ,A>::mapped_type & HashMap<Key,T,H,EQ,A>::operator [](const HashMap<Key,T,H,EQ,A>::key_type& k){ 
    size_type i = hash(k)%b.size(); 

    for(Entry* p=b[i]; p; p=p->next){ 
     if(eq(k, p->key)){ 
      if(p->erased){ 
       p->erased = false; 
       no_of_erased--; 
       return p->val=default_value; 
      } 
      return p->val; 
     } 
    } 

    if(size_type(b.size()*max_load) <= v.size()){ 
     resize(b.size()*grow); 
     return operator[](k); 
    } 

    v.push_back(Entry(k,default_value,b[i])); 
    b[i] = &v.back(); 

    return b[i]->val; 
} 

template<class Key, class T, class H=Hash<Key>, class EQ=equal_to<Key>, class A=allocator<pair<const Key,T>>> 
void HashMap<Key,T,H,EQ,A>::resize(size_type s){ 
    size_type i = v.size(); 
    while(no_of_erased){ 
     if(v[--i].erased){ 
      v.erase(&v[i]); 
      --no_of_erased; 
     } 
    } 

    if(s<=b.size()) 
     return; 
    b.resize(s); 
    fill(b.begin(),b.end(),0); 
    v.reserve(s*max_load); 

    for(size_type i=0;i<v.size(); i++){ 
     size_type ii = hash(v[i].key)%b.size(); 
     v[i].next = b[ii]; 
     b[ii] = &v[i]; 
    } 
} 

template<class Key, class T, class H=Hash<Key>, class EQ=equal_to<Key>, class A=allocator<pair<const Key,T>>> 
void HashMap<Key,T,H,EQ,A>::erase(iterator p){ 
     if(p->erased == false) no_of_erased++; 
     p->erased = true; 
} 


int main(){ 
    return 0; 
} 

Answer 1

變化

HashMap<Key,T,H,EQ,A>::mapped_type & HashMap<Key,T,H,EQ,A>::operator [] 

到

typename HashMap<Key,T,H,EQ,A>::mapped_type & HashMap<Key,T,H,EQ,A>::operator [] 

因爲您的錯誤已經提出。編譯器無法自行推斷mapped_type是類模板中的typedef。