C ++成员函数名为begin（）问题

Question

Assuming that:

vector<string> mvec; 

has some elements on it

Partial code:

for(vector<string>::iterator it1 = mvec.begin(); it1 != mvec.end(); ++it1) {  
for(string::iterator it2 = it1->begin(); it2 != it1->end(); ++it2)

since:

it1->begin() deference the object and then invoke the member function begin() of that object, which object is it1 pointing to?

String?

Answer 1

it1指向存储在向量中的字符串类的实例。

Answer 2

由于it1是vector<string>的iterator ，当你解除它时，你会得到一个string 。

Answer 3

This code:

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

int main() {
    vector <string> mvec;
    mvec.push_back("foo");
    mvec.push_back("bar");
    for(vector<string>::iterator it1 = mvec.begin(); it1 != mvec.end(); ++it1) {  
        for(string::iterator it2 = it1->begin(); it2 != it1->end(); ++it2) {
            cout << * it2 << endl;
        }
    }
}

Prints:

f
o
o
b
a
r

The first iterator gives you access to the strings in the vector, and the second to the characters in the strings.

Answer 4

是的，它是string ，它也是一个容器，所以取消引用it2会得到它的个别字符。

Answer 5

Like others have said, dereferencing it1 will get you a string . It may seem weird, given that it1 isn't a pointer, but the iterator class overloads the operator-> , in which case all bets are off in regards to what will happen.