Question

See this code. I need to change the value of a specific element in a 2D string vector using the iterator. I can use for loop with an index to do this. but here what i need is directly use the iterator to refer the element. something like (*ite)[0] = "new name"
Any idea? I added full working code here for your convenience

#include <iostream>
#include <vector>
#include <string>
#include <sstream>

using namespace std;


string convertInt(int number)
{
   stringstream ss;
   ss << number;
   return ss.str();
}

int main()
{
    vector<vector<string>> studentList;
    for(int i=0; i<10; i++){
        vector<string> student;
        student.push_back("name-"+convertInt(i));
        student.push_back(convertInt(i+10));
        studentList.push_back(student);
    }
    vector<string> temp;
    for(vector<vector<string>>::iterator ite = studentList.begin(); ite != studentList.end(); ++ite){
        temp = *ite;
        if(temp[0].compare("name-5")==0){
            cout << "Changeing the name of student 5" << endl;
            // I need to change the studentList[5][0] (original value not the one in temp vector) at here using iterator ite
        }
    }
return 0;
}

Answer 1

Because temp = *ite makes a copy of vector(student), if you modify on temp, it’s not modified on real studentList, that’s why you need (*ite)[0] = "new name" to change value on real element.

Using for loop is a bit “ugly“, use std::find_if instead of for loop:

bool IsFindFifthName(const std::vector<std::string>& student)
{
   return student[0] == "name-5";
}

 std::vector<std::vector<std::string>>::iterator iter 
  = std::find_if(studentList.begin(), studentList.end(), IsFindFifthName);

if (iter != studentList.end() )
{
   (*iter)[0] = " new name";       
}

Or use Lambda if C++11 is available:

std::vector<std::vector<std::string>>::iterator iter 
  = std::find_if(studentList.begin(), studentList.end(), 
    [](std::vector<std::string>& student){ return student[0] == "name-5"; });

 if (iter != studentList.end() )
 {
    (*iter)[0] = " new name";       
 }