i know how to find prime numbers of several numbers, does anyone know hoew to find a prime sumber of a single number inputed by the ser using the following function and printing out if the number is prime

int mai()//asks for input
{
int num;
cout << "enter number:";
cin >> num;

call IsPrime;
cout if prime or not
}

bool IsPrime(int n)
// precondition: n >= 0
// postcondition: returns true if n is prime, else returns false
// returns false if precondition is violated
{
if (n < 2) // 1 and 0 aren't prime
{ return false; // treat negative #'s as not prime
}
else if (2 == n) // 2 is only even prime number
{ return true;
}
else if (n % 2 == 0) // even, can't be prime
{ return false;
}
else // number is odd
{ int limit = int(sqrt(n) + 1); // largest divisor to check
int divisor = 3; // initialize to smallest divisor

// invariant: n has no divisors in range [2..divisor)

while (divisor <= limit)
{ if (n % divisor == 0) // n is divisible, not prime
{ return false;
}
divisor += 2; // check next odd number
}
return true; // number must be prime
}
}
any suggestions would greatly be appreciated

Greetings cblue,

> does anyone know hoew to find a prime sumber of a single number inputed by the ser using the following function and printing out if the number is prime
Yes, I do know, its quite possible too. How did this work? Simple, the if statement is used to express decisions. Formally the syntax is: Where the else is optional. The expression is evaluated; if it is true (if has a non-zero value), statement1, is executed. If it is false, (if has value of zero) and if there is an else part, statement2 is executed instead.

Since an if simply tests the numeric value of an expression, certain coding shortcuts are possible: Instead of: If you have further questions, please feel free to ask. Hope this helps.

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