Hi all i'm new, I started learning a week or so ago and have joined to the forum so I can get some help and eventually contribute to the community. I am reading a number of books and online guides and have come across a problem I cannot solve when creating a simple coin flip randomizer.

#include <ctime>
#include <cstdlib>
#include <iostream>

using namespace std;


int choice;

int randRange (int low, int high)
{
        return rand() % ( high - low ) + low;
    }

int main ()
{

 cout << "flip a coin? \n\n1: Yes\n\n2: No\n";
 cin >> choice;
 if (choice == 1)
    {

            srand ( time( NULL ) );
            randRange( 1 , 10 );
            if (randRange >= 5)
            {
                cout << "Heads";
            }
else
            {
            cout << "Tails";
            }
    }
else if (choice == 2)
            {
            cout << "Boooo";
            return 0;
            }

}

I am getting a error on the if (randRange >= 5) line, the compiler says that c++ forbids a comparison between pointer and integer, but as far as I am aware I am not using a pointer.

reply's would be gratefully welcomed.

You are using 'randRange' as if it were an 'int' variable. However, 'randRange' is a function so in this context it will return a function pointer. A function pointer is an address in memory where the code for a function resides. Essentially your initial attempt is comparing code to a data type. CaptainKraft's solution will fix that particular compiler issue because you are now comparing against the result of performing that function, which produces an int and comparing it to an int type.

#include <ctime>
#include <cstdlib>
#include <iostream>

using namespace std;


int choice;
int randRange (int low, int high)

{
        return rand() % ( high - low ) + low;
    }

int main ()
{

 cout << "flip a coin? \n\n1: Yes\n\n2: No\n";
 cin >> choice;

 while ( choice != 1 || choice != 2)
 {
     cout << "please enter 1 or 2";
     cin >> choice;
 }
 if (choice == 1)
    {
            srand ( time( NULL ) );
            randRange( 1 , 10 );
            if (randRange( 1,10) >= 5)
            {
                cout << "Heads";
            }
else
            {
            cout << "Tails";
            }
    }
else if (choice == 2)
            {
            cout << "Boooo";
            return 0;
            }

}

Thanks very much for the quick reply, a quick follow up question. I have been struggling with while loops for the last few days, the above script is my current attempt to loop the script if the user input is not one of the acceptable values, from what i have read it would seem that a do-while loop would apply but i should be able to do it with a while statement?

currently the program runs and gets stuck in a infinite loop if i enter an invalid input. i thought the prompt to enter an input would solve this, what am i doing wrong :)?

Your loop continues as long as you don't enter 1 *or* you don't enter 2. To visualize that, here's a table:

1  |   2   | !=1  |  !=2  |  Loop?
-----------------------------------
Y  |   N   |   N  |   Y   |  Yes
N  |   Y   |   Y  |   N   |  Yes
N  |   N   |   Y  |   Y   |  Yes

So in other words, you will always loop forever, no matter what.

Quite helpful.

I find if you have trouble with testing the negative condition in your head, do this:

Determine which is easier to state in your mind. In this case, "I want to exit the loop when choice is 1 or 2". Then just negate this logic for the condition of staying in this loop:

"I want to stay in the loop when !(choice == 1 || choice == 2)".

You can also apply DeMorgan's rule to alter the check. DeMorgan's rule is:

!(A && B) = !A || !B
!(A || B) = !A && !B

Thus the condition can also appear as:

!(choice == 1 || choice == 2) -> !(choice == 1) && !(choice == 2)

I am having trouble comprehending this. Currently I think I am asking: When the user input is not either 1 or 2 then ask the question again and retrieve another response, if I am not asking this, how would it be asked?

thanks