Dos program quits

Yah, so I'm trying to learn C++, I go to the first tutorial with cout , that displays simple text. When I test it in Borland, it opens and closes really fast, this is really annoying.... how do i fix this?

Also, is their any hope for people who aren't super smart with computers to learn C++ ? .. And approximately how long does it take to learn for average people, thats the basics of C++ ..
Thanks

This is in the FAQ.. but basically you need to pause the program before it completes execution. Simplest way is to ask the user for input. This will pause the program until you enter something... like such.

Code:

---

#include <iostream>

using namespace std;

int main(void)
{
        cout << "If I don't pause by waiting for input, I will close!" << endl;

        cin.get();  // Ask the user for input

        return 0;
}

---

You can use system("pause"); too

Code:

---

#include <iostream>
using namespace std;

int main()                // int main() never returns a void, main returns always int
{
        cout << "YOUR OUTPUT" << endl;

        system("pause");
       
        return 0;
}

---

I'm sure, that there is a keyboard combination like Ctrl + F5 (Visual Studio) to let the window open for your C++ IDE

when u create a consol app in Borland or Visual c++ u can hold the scrren by asking the user to enter a char or something.

in borland --- getchar();

this function hold ur screen and u can exit the screen by pressing Enter key.

try using the getch function. works for me. just add it right before your return statement and make sure to add the conio.h header file

Code:

---

#INCLUDE <iostream.h>
#include <conio.h>// for getch()

int main()

{

cout << "My screen wont close no more";
getch(); //this waits for a user to entre any key before it closes
return 0;
}

---

best is system("PAUSE");
it gives you the press any key to continue message

Quote:

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Originally posted by Iamien
best is system("PAUSE");
it gives you the press any key to continue message

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You could just type that in. It's only one more line of code.