Writing a program with more than one parameter.

[xcrossmyheartx]

This entire problem has my mind completely in circles.

Here's the original problem:

I tried writing out the program using the logic I thought would work and here's my code:

#include <iostream>
#include <string>

using namespace std;
double displayDivision(double x,double y)
{
	double answer;
	answer = (x / y);
	// if (divisor == 0)
	  // {
	      //cout << "Error: Attempt to divide by zero!" << endl;
	  // }
	// else 
	  // 

}

int main ()
{
	// put this is in a do..while function?
	double dividend, divisor;
	cout << "Enter the dividend: ";
	cin >> dividend;
	cout << "Enter the divisor: ";
	cin >> divisor;
	// call function
	// cout << "Do you want to continue (y/n)?
	// cin >> response;
	// if (response == "n") || (response == "N")
	//		// return 0;

}

My first problem that I run into is that I'm supposed to have my function display the division and not return it. This means that instead of return; it's going to simply be a cout << right?

When I try to run this, I get an error that says my function must return a value.

Any ideas? Maybe my logic is really flawed.

[Agnimukha]

Line 5 tells the caller to expect a double to be returned. If you put void in there you don't have to return anything.

[R0mai]

This means that instead of return; it's going to simply be a cout << right?

Yes

When I try to run this, I get an error that says my function must return a value.

If you don't want a function to return anything use a void function.

Check for zero divisor before doing the calculation.
You will need to use a loop in the main function, which runs until the user wants to continue (I recommend a do-while loop, since you don't wanna ask the user if he/she want's to continue in the first iteration). In the loop you first ask for user input, call your function, then ask the user if he/she want's to continue.

Good luck, come back if you have any further troubles.

[xcrossmyheartx]

Here's what I have so far. For some reason, I keep getting that choice is undeclared. I'm clearly defining it, aren't I?

#include <iostream>
#include <string>

using namespace std;
void displayDivision()
{
	
}

int main ()
{
	do
	{
	double dividend, divisor;
	

	cout << "Enter the dividend: ";
	cin >> dividend;
	cout << "Enter the divisor: ";
	cin >> divisor;

	if (divisor == 0)
	{
		cout << "Error: Attempt to divide by zero!" << endl;
	}
	else 
	{
		double response = (dividend / divisor);
		displayDivision();
	}
    char choice;
	cout << "Do you want to continue (y/n)? ";
	cin >> choice;
	} while ((choice == 'y') || (choice == 'Y'));
		
	
	
	

}

[Zhuge]

The condition of your while loop can't see choice declared inside of the loop.

[Galik]

The while condition happens outside the loop:
while ((choice == 'y') || (choice == 'Y'));
So you need to declare choice outside the loop also.

[xcrossmyheartx]

Okay.. I moved it before the loop starts and is letting me run. However, the loop isn't running properly.

I tried entering a y and it just ended.

[xcrossmyheartx]

I'm baffled, I don't know why my loop isn't working. I just rewrote it just to make sure I didn't typo and I can't see anything wrong with it.

#include <iostream>
#include <string>

using namespace std;

void displayDivision()
{
	cout << "hi" << endl;
}


int main ()
{
	double dividend, divisor, choice;

	do
	{
	cout << "Enter the dividend: ";
	cin >> dividend;

	cout << "Enter the divisor: ";
	cin >> divisor;

	if (divisor == 0)
	{
		cout << "Error: Attempt to divide by zero!" << endl;
	}

	else 
	{
		double response = (dividend / divisor);
		cout << response << endl;
		displayDivision();
	}

    cout << "Do you want to continue (y/n)? ";
	cin >> choice;
	
	} while ((choice == 'y') || (choice == 'Y'));
		
	
	
	

}

[Galik]

double dividend, divisor, choice;

Isn't choice meant to be a char?

[xcrossmyheartx]

Oh wow ~_~ One little thing can throw the whole code off.

I feel dumb. Thanks for the response, lol.

The last thing I have to do is make the function. I'm really unsure of how to do this since I can't reference anything.

I know I have to cout << something but I can't do cout << response

Any ideas?

[Galik]

http://www.cplusplus.com/doc/tutorial/basic_io/

[xcrossmyheartx]

Hm, didn't get anything from that..

I am trying to make my function,

void displayDivision()
{
	cout << "hi" << endl;
}

Output the answer without storing it.

Am I missing something big in the link you posted?

[xcrossmyheartx]

I don't understand how this is supposed to work if the function and the program are two different things >________> urgh, mind bend.

[Galik]

You were pretty close with your original code for this function. The code you commented out looked good.

[xcrossmyheartx]

\assignment27.cpp(19) : error C4716: 'displayDivision' : must return a value

#include <iostream>
#include <string>

using namespace std;

double displayDivision(double dend, double visor)
{
	double answer = (dend / visor);
	if (visor == 0)
	{
		cout << "Error: Attempt to divide by zero!" << endl;
	}

	else 
	{
		cout << answer << endl;
		}
		
}


int main ()
{
	double dividend, divisor;
	char choice;

	do
	{
	cout << "Enter the dividend: ";
	cin >> dividend;
	


	cout << "Enter the divisor: ";
	cin >> divisor;

	//call the function
	


	

    cout << "Do you want to continue (y/n)? ";
	cin >> choice;

	} while ((choice == 'y') || (choice == 'Y'));
	
	
	
	

}

The instructions tell me not to call the function. /sigh/

[Galik]

Why? Just use an if() statement in your function to see if the divisor is zero.

You do know what the divisor is don't you?

[Galik]

Read up on this:

http://www.cplusplus.com/doc/tutorial/functions/

It might help.

[xcrossmyheartx]

I've read that at least a dozen times as it's been suggested before. I know what a function does and how to reference them in the examples they give. The problem says that I have to print this in the function without returning the information.

And yes, the divisor is what I'm dividing by.

[Galik]

Then you have all the pieces. Like I said, your original code that you commented out was fairly close.

[m4ster r0shi]

xcrossmyheartx wrote:

The instructions tell me not to call the function.

Where do they tell that?

xcrossmyheartx's compiler said:

\assignment27.cpp(19) : error C4716: 'displayDivision' : must return a value

As R0mai said, use a void function. A void function is a function that doesn't return a value. It doesn't mean that it can't take any arguments. That's how your function's signature should look like:

void displayDivision(double dividend, double divisor)