在C ++中映射内部结构

Question

I defined a structure and a vector of the structure.I took input by a temporary structure of the type bind and pushed it back into the vector.

typedef struct {
        string bindername ;
        map<string ,long long int> m ;
        int index ;
    }bind ; 
    vector <bind> v ;

Inside main function , I declared an iterator it as :

map <string ,long long int>::iterator it ;

When I am trying to access vector's member by iterator it , I am having errors . I tried to access it like this :

int l = v.size () ;
for (int K = 0 ; K < l ; K++)
        {
            for (it = v[K].m.begin () ;it != v[K].m.end () ; it++) // this line is generating errors
            {
                if ((it->second ()) < b) //this line is also generating errors // b is an int
                {
                    cout << it -> first () << endl ;
                    c++ ;
                }
            } 

Errors generated :

b.cpp: In function ‘int main()’:
b.cpp:84:22: error: ‘it.std::_Rb_tree_iterator<_Tp>::operator-> [with _Tp = std::pair<const std::basic_string<char>, long long int>, std::_Rb_tree_iterator<_Tp>::pointer = std::pair<const std::basic_string<char>, long long int>*]()->std::pair<const std::basic_string<char>, long long int>::second’ cannot be used as a function
b.cpp:86:27: error: no match for call to ‘(const std::basic_string<char>) ()’

How can I access V[K].m's second elements by iterator it to check whether it is less then b or not ?

Answer 1

You do it->second () which is a function call, but the second member is not a function, it's an integer. You do the same with the first element of the iterator two lines down.

Answer 2

The two public members of std::pair are objects, not functions.

So it->second() should be it->second .

Answer 3

if ((it->second ()) < b)
//              ^^

You have a set of parenthesis so the compiler thinks you're incorrectly calling a function.

Answer 4

it->second < b

第二个是数据成员，而不是函数。