CODEWITHSUNDEEP
c++pointersvector
Faizan Karim
Faizan Karim

Reputation: 38

c++ program printing out address instead of value

I have a program that takes in information through a struct and puts it into a vector, and I'm trying to print that information out but instead get an address. The structure should hold the values correctly so I think it's either my pointers or the way I'm printing it out.

#include <iostream>
#include <cstring>
#include<vector>

using namespace std;

struct student
{
  char* fName;
  char* lName;
  int id;
  float gpa;
};

void add(vector<student*>*);

int main() 
{
  vector <student*>* list = new vector<student*>();


    if (strcmp(cmd,"ADD") == 0)
    {
      add(list);  
    }
    else if (strcmp(cmd,"PRINT") == 0)
    {
      for(vector<student*>::iterator i = list->begin(); i != list->end(); i++)
      {
        cout << *i;
      }
      cout << "print" << endl;
    }
}
  void add(vector<student*>* paramlist)
  {
    student* s = new student();
    s->fName = new char[25];
    s->lName = new char[25];

    cout << "Enter first name" << endl;
    cin >> s->fName;

    cout << "Enter last name" << endl;
    cin >> s->lName;

    cout << "Enter id number" << endl;
    cin >> s->id;

    cout << "Enter GPA" << endl;
    cin >> s->gpa;

    paramlist->push_back(s);
  }

Or it might have something to do with the way I iterate through the vector.

Upvotes: 1

Views: 980

Answers (2)

Karl Nicoll
Karl Nicoll

Reputation: 16419

You need to add an operator overload for your struct, to define how the struct should appear when printed. You also need to dereference the pointer as well as the iterator.

// Define how the struct should look when printed.
// This function makes it appear like:
// Name: John Smith, ID: 1235, GPA: 4.0
std::ostream &operator<<(std::ostream &os, const student &val) {
  os
    << "Name: " << val.fname << " " << val.lname
    << ", ID: " << val.id
    << ", GPA: " << val.gpa
    << endl;

  return os;
}

Then later...

for(vector<student*>::iterator i = list->begin(); i != list->end(); i++)
{
  // Dereference twice, once for the iterator, and again for the pointer.
  cout << **i << endl;
}

Upvotes: 1

Askold Ilvento
Askold Ilvento

Reputation: 1510

You have to dereference twice **i. With *i you get the address of vector<student*> element that is student*. You get student, you need other *.

You may use for (auto i: list) to make your life easier.

Upvotes: 0

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