Thread: Glass Rod dilemma.

Good Morning all:

I am new to programming and I have a homework assignment that is giving me some difficulty. The problem is the glass rod problem. Where you break a glass rod in 3 pieces an see if the pieces will for a triangle.
The main problem is how the professor wants the program written.
We have to write a main and three functions that do most of the work. I have my code written and it compiles but it returns erroneous answers. Here is my code, can anyone point me in the right direction to fixing it. Please

Code:

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


using namespace std;


void doBreak (int & side1, int & side2, int & side3);
bool attempts (int);
void randomNumberGen (int & num1, int & num2);


void main()
{
    int count = 1;
    int tests = 1000000;
    double probability;
    int attempts;
    int triangle = 0;
    int val;
    
    srand (time(0));


    do
        {
            triangle = triangle + attempts;
            count++;
        }while (count <= tests);


    probability = (triangle / tests) * 100;


    cout <<"The probability that the broken glass rod will form a triangle is"
        << probability <<"% \n";
}


bool attempts (int)
{
    int side1, side2, side3;


    doBreak (side1, side2, side3);
    if 
        ((side1 + side2) > side3 &&
        (side1 + side3) > side2 &&
        (side2 + side3) > side1)
    {
        return true;
    }
    else
    {
        return false;
    }
}


void doBreak (int & side1, int & side2, int & side3) 
{
    int num1, num2;
    randomNumberGen (num1, num2);


    side1 = num1;
    side2 = (num2 - num1);
    side3 = 1000 - num2;
}


void randomNumberGen (int & num1, int & num2)
{
    do
  {
    num1 = rand() % 999 + 1;
    num2 = rand() % 999 + 1;
  } while (!(num1 < num2));
}

Hello Salem:
If I am reading your response correctly, ( triangle = triangle + attempts is line 30 of the code. do I need to add the (>) symbol?

I have tried ( attempts ();, attempts (int);, attempts (int, int, int) It always says that " term does not evaluate to a function taking however many arguments. What am I doing wrong?

I made the following changes and now my answer comes back 0% every time.

Code:

do
		{
			attempts();
			triangle = triangle + attempts();
			count++;
		}while (count <= tests);


	probability = (triangle / tests) * 100;


	cout <<"The probability that the broken glass rod will form a triangle is"
		<< probability <<"% \n";
}


bool attempts ()
{
	int side1, side2, side3, triangle;


	doBreak (side1, side2, side3);
	if 
		((side1 + side2) > side3 &&
		(side1 + side3) > side2 &&
		(side2 + side3) > side1)
	{
		return  triangle = true;
	}
	else
	{
		return  triangle = false;
	}
}

Thank you Salem. I made the change you suggested plus a couple more and now it works great thank you very much. Here is the working code.

Code:

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iomanip>


using namespace std;


void doBreak (int & side1, int & side2, int & side3);
bool attempts ();
void randomNumberGen (int & num1, int & num2);


void main()
{
	int count = 1;
	double tests = 1000000;
	double probability;
	double triangle = 0;
	
	
	srand (time(0));
	
	do
		{
			attempts();
			triangle = triangle + attempts();
			count++;
		}while (count <= tests);


	probability = ((triangle * 100) / tests);


	cout << setprecision(5); 
	cout <<"The probability that the broken glass rod will form a triangle is"
		<< probability <<"% \n";
	
}


bool attempts ()
{
	int side1, side2, side3, triangle;


	doBreak (side1, side2, side3);
	if 
		((side1 + side2) > side3 &&
		(side1 + side3) > side2 &&
		(side2 + side3) > side1)
	{
		return true;
	}
	else
	{
		return false;
	}
}


void doBreak (int & side1, int & side2, int & side3) 
{
	int num1, num2;
	randomNumberGen (num1, num2);


	side1 = num1;
	side2 = (num2 - num1);
	side3 = 1000 - num2;
}


void randomNumberGen (int & num1, int & num2)
{
	do
  {
    num1 = rand() % 999 + 1;
    num2 = rand() % 999 + 1;
  } while (!(num1 < num2));
}