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coin flip problem


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#1 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 09:24 AM

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.



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#2 CaptainKraft   Members   -  Reputation: 265

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Posted 11 September 2013 - 09:31 AM

My experience with C++ is pretty limited but I think I might be able to help here.

 

It looks like you are calling the randRange function before entering the if statement, and you are assuming that the result is stored in a variable that is called randRange. What you need to do is either store the return value of the function call into a variable, or just call the function within the if statement. I prefer the latter, and here's a small example:

if (randRange(1, 10) >= 5) {
    // do stuff here
}

That should solve your current problem.



#3 Cosmic314   Members   -  Reputation: 1147

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Posted 11 September 2013 - 09:45 AM

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.



#4 Glass_Knife   Moderators   -  Reputation: 3434

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Posted 11 September 2013 - 09:46 AM

The randRange() is a function.  You have to assign the result of the function to a new variable, and try that.  (just using the function name the compiler thinks you're using a function pointer.  Don't worry about that.  Try something like this:


int randomNumber = randRange( 1 , 10 );
if (randomNumber >= 5) { ... more code here }

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#5 Matias Goldberg   Crossbones+   -  Reputation: 3007

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Posted 11 September 2013 - 09:48 AM

Hi!
 
The problem is that "randRange" is the name of a function. So are you comparing a function against the number 5 (cows vs apples)
 
Like CaptainKraft said, changing it to
"if (randRange(1, 10) >= 5)" will execute the function randRange, and compare its return value with the number 5. Since you're comparing numbers with numbers, it will work.
 
Note this pattern is often considered hard to read, so it's clearer if you assign the return value to a temporary variable and then do the comparison:
int flipCoinResult = randRange( 1 , 10 );
if (flipCoinResult >= 5)
{
   cout << "Heads";
}


#6 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 11:43 AM

#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 :)?



#7 Washu   Senior Moderators   -  Reputation: 4469

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Posted 11 September 2013 - 11:54 AM

As for the invalid input, check if the input stream is still valid. If its not you have a few choices, one of which would be to simply flush the stream.

 

Read up on std::basic_ios::clear, std::basic_ios::fail, and std::basic_istream::ignore.

 

Additionally, choice should not be a global.


Edited by Washu, 11 September 2013 - 11:58 AM.

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#8 ApochPiQ   Moderators   -  Reputation: 14282

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Posted 11 September 2013 - 12:02 PM

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.

#9 Cosmic314   Members   -  Reputation: 1147

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Posted 11 September 2013 - 12:18 PM

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)


#10 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 12:46 PM

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



#11 SiCrane   Moderators   -  Reputation: 9387

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Posted 11 September 2013 - 12:51 PM

Not either 1 or 2 would be !(choice == 1 || choice ==2).



#12 IndyOfComo   Members   -  Reputation: 522

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Posted 11 September 2013 - 12:56 PM

 

 

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?

 

Also, Mr Stx, I would highly suggest tackling one error at a time and not adding code when you know you have an error. This practice should make it easier to fix things as you find them. Above, you have changed the main line to loop, but you also still have the problem of treating randRange() as a variable. Now, this error has nothing to do with your infinite loop--the replies you've gotten on that should be helpful--but that just happens to be the case in this situation. If you add too much all at one time, it can become much more difficult (as a beginner) to figure out where the problem is really coming from. So fix the randRange() thing, then straighten out your boolean algebra(woot!), and then move on.

 

While I was typing this I see you have another post. I put in a link above, I don't know how helpful it will be.


Here is my technical background info.

#13 IndyOfComo   Members   -  Reputation: 522

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Posted 11 September 2013 - 01:01 PM

Not either 1 or 2 would be !(choice == 1 || choice ==2).

 

Take a closer look at how you wrote your code, and answer each question for yourself in each situation.

while ( choice != 1 || choice != 2)

input     test 1   --   test 2   --   summary

  1           F               T              (F || T) = T

  2           T               F              (T || F) = T

  3           T               T              (T || T) = T

 

Step through SiCrane's code and see what you get in each of these cases.


Here is my technical background info.

#14 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 01:08 PM

Thanks guys ill spend a few hours looking into the information you have kindly supplied me and if I'm still stuck ill post again.



#15 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 02:50 PM

ok I have done a good bit of research and have the information provided in this thread great starting points, I have come across a hurdle regarding the "!(choice == 1 || choice ==2)" statement, I understand the concept of it but the compiler doesn't allow me to enter this as the "!" isn't in brackets.



#16 SiCrane   Moderators   -  Reputation: 9387

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Posted 11 September 2013 - 02:55 PM

You would wrap the whole thing in parenthesis. I.e.:
if (!(choice == 1 || choice == 2))


#17 Paradigm Shifter   Crossbones+   -  Reputation: 5126

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Posted 11 September 2013 - 02:56 PM

if(!(choice == 1 || choice ==2))

 

should work.

 

Are you doing

 

if !(choice == 1 || choice ==2)

 

that won't work, "if" needs brackets straight away.


Edited by Paradigm Shifter, 11 September 2013 - 02:57 PM.

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#18 Mr Stx   Members   -  Reputation: 116

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Posted 11 September 2013 - 04:15 PM

 while (!( choice == 1 || choice == 2))

 

still gives me the infinite loop on an invalid input



#19 Servant of the Lord   Crossbones+   -  Reputation: 17125

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Posted 11 September 2013 - 04:38 PM

Could you post the current state of your code?


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#20 SiCrane   Moderators   -  Reputation: 9387

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Posted 11 September 2013 - 04:39 PM


 while (!( choice == 1 || choice == 2))

 

still gives me the infinite loop on an invalid input

You mean on things that aren't integers? That's when you go back up to Washu's post and read about fail() and clear(). Short version: if a stream contains something that it wasn't expecting, like a letter when you're trying to extract an int, the stream sets the fail bit and causes all subsequent reads to fail until you handle the error. Personally, rather than read integers directly from the stream, I prefer to read lines into a string with std::getline() and dump those strings into a stringstream and extract from that.






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