Worked on this C++ function assignment for about 8 hours!!! Please post solution.

[2kaud]

Humm.. It's not very efficient. As prime numbers cannot be even, why test every even number in the loop? You can just test if divisible by 2 at the beginning and then start the loop at 3 and increment by 2. Also, there's no need to test all numbers up to the one being tested. As a number squared is divisible by the number, the loop test can stop at the square root of the tested number.

//A Prime Number can be divided evenly only by 1, or itself
//So if a number(n) can be divided by any number between 2 and n-1 inclusive then it is not prime
//An even number is never prime
bool IsPrime(int num)
{
	//numbers 1 and 2 are defined as prime
	if ((num == 1) || (num == 2))
		return true;

	//Even numbers are never prime
	if ((num % 2) == 0)
		return false;

const int limit = (const int)sqrt((double)num);

	//Need only consider odd numbers before sqrt
	for (int i = 3; i <= limit; i += 2)
		if ((num % i) == 0)
			return false; // not prime

	return true;
}

Why not use the IsPrime function in GetPrimes rather than repeat the same code twice? Also passing the vector by reference is more efficient than returning a vector.

typedef vector<int> PRIMES;

//Put the first N prime numbers into a vector
void GetPrimes(PRIMES& vp, int no)
{
int	prime = 3,	//Start primes at 3
	fnd = 0;	//Number found

	while (fnd <= no) {
		if (IsPrime(prime)) {
			vp.push_back(prime);	//Save prime
			fnd++;
		}
		prime +=2;	//Only odd numbers
	}
	return;
}

Again, with ReverseInteger, there is a much more efficient method without the overhead of converting to and from a string

//Reverse an integer
int ReverseInteger(int no)
{
int rnum = 0;

	while (no) {
		rnum = rnum * 10 + (no % 10);
		no /= 10;
	}

	return rnum;
}

So the main program now looks like

int main()
{
int	num,
	prime,
	rprime;

PRIMES	Vp;

	cout << "Enter number of primes to check for emirp status: ";
	cin >> num;

	GetPrimes(Vp, num);

const size_t noprimes = Vp.size();

	for (size_t i = 0; i < noprimes; i++) {
		prime = Vp[i]; 
		rprime = ReverseInteger(prime);
		cout << prime << " is prime ";
		if (IsPrime(rprime)) {
			cout << "and " << rprime << " is prime so " << prime << " is emirp " << endl;
		} else {
			cout << "but " << rprime << " is not prime so " << prime << " is not emirp " << endl;
		}
	}

	return 0;
}