Why am I getting warnings that function "doesn't always return a value" ?

I have been working on this program that is a timed mathematical quiz and have everything completed. The problem is I keep getting warnings when I compile that says 1: 'sign' : not all control paths return a value and 2: 'equation' : not all control paths return a value. Can anyone help me see what the problem is?

Code:

 
#include<iostream>

using std::cout;
using std::cin;
using std::endl;

#include<cstdlib>

#include<ctime>

int num();                                // function prototype 
char sign();                              // function prototype
int equation( int a, char c, int b );     // function prototype

int main()
{
	srand( time( 0 ) );   // This time function returns the current 
	                      // "calendar time" in seconds.  This value is
	                      // converted to an unsigned integer and used as 
	                      // the seed to the random number generator.

	int numcorrect = 0;  
	int stop = time( 0 ) + 30;
	int x, y, answer, correctanswer;
	char z;

	while ( time( 0 ) < stop )
	{	
		x = num(), y = num();    // Assignment of x and y to the num function
		z = sign();

		cout << "What is " << x << " " << z << " " 
			 << y << endl;       // Prompt to ask the arithmetic question
		cin >> answer;           // Input answer

		correctanswer = equation( x, z, y ); // Assignment of correctanswer
		                                     // to the equation function

		if ( correctanswer == answer )       // tests user's answer
		{
			cout << "Very good!\n\n"; 
			++numcorrect;
		}
		else
		{
			cout << "Incorrect.\n"; 
			cout << "The correct answer is " << correctanswer << "\n\n";
		}

	}                                        // end while loop

	cout << "\nYou answered " << numcorrect 
		 << " questions correctly.\n\n";     // Prompts # of correct answers
	
	return 0;
}

// Function implementation
int num() 
{
	return 1 + ( rand() % 12 );    // returns random integer between 1 and 12
}

// Function implementation
char sign()
{
	int k = 1 + rand() % 5;   // assigns random integers between 1 and 5 to k

	if ( k == 1 )             // tests k and returns the + sign
		return '+';

	if ( k == 2 )             // tests k and returns the - sign
		return '-';

	if ( k == 3 )             // tests k and returns the * sign
		return '*';

	if ( k == 4 )             // tests k and returns the / sign
		return '/';

	if ( k == 5 )             // tests k and returns the % sign
		return '%';
}                             // ends the sign function implementation block

// Function implementation
// a, c, and b below are parameters to 
// the equation function implementation
int equation ( int a, char c, int b )
{
	if ( c == '+' )           // tests c to return the appropriate equation
		return a + b;

	if ( c == '-' )
		return a - b;

	if ( c == '*' )
		return a * b;

	if ( c == '/' )
		return a / b;

	if ( c == '%' )
		return a % b;
}                            // ends equation function implementation block

Thank you!
teeCee

There are no else's in those functions, so if something doesn't match any of the if statements, it falls off the end of the function without returning anything.

Thanks alot!!

Wouldn't the else statements really be unnecessary though since the way it's set up, there are no possibilities for a false?

I wanted to use the case structure instead, but I just wasn't able to get it to work.

Thanks again,
teeCee

If you really want to shut the compiler up, just stick a return 0; at the end of both of the functions. In some instances this warning is useful, others, it's just annoying.

In your situation, its unnecessary. That's why its a warning and not an error. The compiler doesn't exactly know what is going to happen, but it is possible for your function to not return anything, as it sees it.