Pointer arrays, oo, fun! >.< URG.

Code:

---


#include <iostream.h>
#include <conio>
using namespace std;

int main(){
        char array[10];

        for (int i = 0; i <= 9; i++){
                cout << "Line " << i + 1 << ": ";
                cin.getline(*(array + i));
                //while(cin.get() != '\n'){}
                //cin.ignore();
                //getch();
                cout << endl;
        }
       
        cout << endl;
       
        for (int i = 0; i <= 9; i++)
                cout << "Line " << i + 1 << "contains string: " << *(array + i) << endl;
       

    return 0;
} //end int main

---

PROBLEM:
Write a function that accepts ten lines of user-input text and stores the entered lines as ten individual strings. Use a pointer array in your function.

REAL PROBLEM: I can't clear the input buffer correctly. I've tried everything. Cin.getline won't work either

I am also not sure if I am using a pointer array

:confused:
No offence friend, but I think you're getting a little ahead of yourself! You need to master the basics of this language first.
:confused:

try this.

Code:

---

#include <iostream.h>

#define PTR_ARR_LEN 10
#define STR_LEN 20

int main()
{
  char *Array[PTR_ARR_LEN];

  for (int i = 0; i <= PTR_ARR_LEN - 1; i++)
  {
    Array[i] = new char[STR_LEN];
    cout << "Line " << i + 1 << ": ";
    cin.getline(Array[i], STR_LEN);
    cout << endl;
  }

  for (int i = 0; i < PTR_ARR_LEN; i++)
    cout << "Line " << i + 1 << " contains string: " << Array[i] << endl;

  getchar();

  return 0;
}

---

Now see, this just makes no sense. My code had similar syntax in that foor loop earlier but the input buffer would not budge...

Does this have to do with the = new char[] thing? What's that do?

Before you had just declared an array of char pointers, but these pointers were not allocated any memory. They were just a load of variables holding random data, so before when you were trying to use cin.getline e.g. entering "hello", and you were putting this into array[0], there was no memory allocated to fit "hello".
Therefore, when you say:
Array[i] = new char[STR_LEN]; I would char pointer number one to hold a size string of 20 chars. Hope this helps a bit.

I think Sebastiani was right.