Erro: Need typename because vector<type> is a dependent scope

[jgg2002]

I am writing a class for a priority vector, but I am facing trouble when I declare an iterator with <type> which is the typename of my class template. The error is in line 37.

error: need 'typename' before 'std::vector<type>::iterator' because 'std::vector<type>' is a dependent scope|

  #include<iostream>
#include<vector>
#include<typeinfo>

using namespace std;

string first (string a, string b){
int siz = min(a.size(), b.size());
int place = 0;
while (a[place]==b[place] && place<siz){
    place++;
}
if (a[place]>b[place]){
    return b;
}
else if (a[place]<b[place]){
    return a;
}
}

template <class type>
class priority_vector{
public:
type check;
string check2;
vector <type> pv;
bool isString(){
if (typeid(type).name()==typeid(check2).name()){return true;}
else {return false;}
}
void push_in(type input){
if (isString()){
    if (pv.size()==0){
        pv.push_back(input);
    }
    else{
        vector <type> :: iterator vit;
        vit = pv.begin();
        while (first(*pv, input)==*pv){
            pv ++;
        }
        vit.insert(vit, input);
    }
}
else{
    if(pv.size()==0){
        pv.push_back(input);
    }
    else{
        vector <type> :: iterator vit;
        vit = pv.begin();
        while (*pv<input){
            pv++;
        }
        vit.insert(vit, input);
    }
}
}
type operator[](int a){
return pv[a];
}
};

int main(){
priority_vector <string> v_s;
int n;
cin>>n;
int in;
for (int i=0; i<n; i++){
    cin>>n;
    v_s.push_in(n);
}
for (int i=0; i<n; i++){
    cout<<v_s[i];
}
return 0;
}

[Peter87]

The error is pretty clear. You need to use typename to tell the compiler that vector<type>::iterator is a type, otherwise the compiler will assume it's a variable or function.

typename vector<type>::iterator vit;

[jgg2002]

Yeah, but why does it need that? Type was already declared at the template.

[JLBorges]

> why does it need that? Type was already declared at the template.

See: http://www.cplusplus.com/forum/beginner/103508/#msg557627

Note: a class template C<T> can be specialised;
consequently C<double>::some_name could be something quite different from C<int>::some_name.