Pointer To Char Array (cin & cout)

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Hello Guys, ..

just to refresh my concept i'm working on parallel arrays. One is used to store integer data and the other one for char data i.e GPA .

The problem compiles like a charm but the result is not correct it displays the Student IDs correctly but not the GPA that's the problem.

- The simple cin works fine
- I don't really know how to use cin.get and cin.getline using pointers. ??
- Please help me in that, Here in enter function I want to get the 2 char long string +1 for null char.
- Code Below

C++

#include <iostream>
#include <cstring>
using namespace std;

void enter(int *ar, char *arr, int size);
void exit(int *a, char *yo, int size);

int main()
{
   const int id = 5;
   const char grade = 5;
    int *student = new int[id];
    char *course = new char[grade];
    cout << "\n";

    enter(student, course, 5);
    exit(student, course, 5);

}

void enter(int *ar, char *arr, int size)
{
    for(int i = 0; i < size; i ++)
    {
        cout << "Student ID: " << i+1 << "\n";
        cin >> *(ar+i);
        cin.ignore();
        cout << "Student Grade: " << i+1 << "\n";
        cin.get(arr, 3);
    }
}

void exit(int *a, char *yo, int size)
{
    for(int i = 0; i < size; i ++)
    {
        cout << "ID And Grade Of Student #" << i+1 << ":";
            cout << *(a+i) << "\t" << *(yo+j) << endl;
    }
}

Version_2 (updated)

C++

#include <iostream>
#include <cstring>
using namespace std;

void get_data(int *ar, char *arr, int size);
void show_data(int *a, char *yo, int size);

int main()
{
   const int id = 5;
   const char grade = 5;
   const int arraySize = 5;
    int *student = new int[id];
    char *course = new char[grade];
    cout << "\n";

    get_data(student, course, arraySize);
    show_data(student, course, arraySize);

	return 0;

}

void get_data(int *ar, char *arr, int size)
{
    for(int i = 0; i < size; i ++)
    {
        cout << "Student ID: " << i+1 << "\n";
        cin >> *(ar+i);
        cin.ignore();
        cout << "Student Grade: " << i+1 << "\n";
        //cin.get(arr, 3);
		cin.get(*(arr+i));
    }
}

void show_data(int *a, char *b, int size)
{
    for(int i = 0; i < size; i ++)
    {
        cout << "ID And Grade Of Student #" << i+1 << ":";
            cout << *(a+i) << "\t" << *(b+i) << endl;
    }
}

Posted 29-Oct-13 8:46am

Usman Hunjra

Updated 29-Oct-13 22:00pm

v3

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Comments

Aescleal 30-Oct-13 4:33am

Why are you using pointers, dynamic memory allocation and raw character arrays when std::vector and std::string would make your life a lot easier? You're writing too much low level code for the scale of the problem.

3 solutions

Solution 3

#include <iostream>
#include <cstring>
using namespace std;

void enter(int *ar, char *arr, int size);
void exit(int *a, char *yo, int size);

int main()
{
const int id = 5;
const char grade = 5;
int *student = new int[id];
char *course = new char[grade];
cout << "\n";

enter(student, course, 5);
exit(student, course, 5);

}

void enter(int *ar, char *arr, int size)
{
for(int i = 0; i < size; i ++)
{
cout << "Student ID: " << i+1 << "\n";
cin >> *(ar+i);
cin.ignore();
cout << "Student Grade: " << i+1 << "\n";
cin.get(arr, 3);
}
}

void exit(int *a, char *yo, int size)
{
for(int i = 0; i < size; i ++)
{
cout << "ID And Grade Of Student #" << i+1 << ":";
cout << *(a+i) << "\t" << *(yo+j) << endl;
}
}

Posted 30-Oct-13 5:10am

Member 10369941

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CPallini · Accepted Answer · 2013-10-29T10:33:00

Solution 1

If the student grade is a single character, then you have to replace

Quote:
cin.get(arr, 3);

with

C++

cin.get(*(arr+i));

Please note

Quote:
cout << *(a+i) << "\t" << *(yo+j) << endl;

you should replace j with i in order to compile.

Posted 29-Oct-13 10:33am

CPallini

Comments

Usman Hunjra 30-Oct-13 3:59am

yes sir, it works like a charm ..
but i actually want to enter more then one characters..
please help me in that ..
PLEASE ...

CPallini 30-Oct-13 4:53am

Why dont you use std::string, then? It works like a charm with std::cin... :-)

nv3 · Accepted Answer · 2013-10-29T11:04:00

CPallini has already shown how to fix your program. Here are a couple more hints about programming style that might help you along your way.

Good naming is very important in a program. In your example, some of the names are misleading:

C++

const int id = 5;
const char grade = 5;

How about naming them arraySize. And you just need one, because by design both arrays should always have the same length. And once we have defined this nice constant, why not use it all places where the array size is specified:

C++

enter(student, course, arraySize);
exit(student, course, arraySize);

By the way, exit is a very misleading name for that function. You certainly meant to call it show_data or something similar.

One of the tricky things in C is that arrays and pointers are somehow related and every C textbook will tell you that the expression "*(ar+i)" is exactly the same as "ar[i]". Only, the latter is a lot more intuitive. So why don't we write:

C++

cin >> ar[i];

and

C++

cout << a[i] << "\t" << yo[i] << endl;

There are still more things to improve; for example the parameter names of the two functions (ar, yo ...) are all but intuitive.

Hope that helps you improve your programs.