C++ Prime Number Program: Check and Print Primes

A prime number is a whole number greater than 1 that can only be divided evenly by 1 and itself. Writing a program to detect primes is one of the best early exercises in C++ because it combines loops, conditionals, and a little math reasoning into a small, satisfying problem.

What Counts as Prime

The numbers 2, 3, 5, 7, 11, and 13 are prime. Numbers like 4, 6, 8, and 9 are not, because they have factors other than 1 and themselves. The numbers 0 and 1 are special cases: neither is prime, so we handle them separately before doing any real work.

The Basic Approach

To check whether a number n is prime, we try to divide it by every integer from 2 up to n - 1. If any division leaves no remainder, n has a factor and is therefore not prime.

#include <iostream>

int main() {
    int n;
    std::cout << "Enter a number: ";
    std::cin >> n;

    bool isPrime = true;

    if (n <= 1) {
        isPrime = false;  // 0 and 1 are not prime
    } else {
        for (int i = 2; i < n; i++) {
            if (n % i == 0) {  // i divides n evenly
                isPrime = false;
                break;         // no need to keep checking
            }
        }
    }

    if (isPrime)
        std::cout << n << " is prime.\n";
    else
        std::cout << n << " is not prime.\n";

    return 0;
}

The % (modulo) operator gives the remainder of a division. When n % i == 0, it means i divides n with nothing left over — a factor. The break statement stops the loop early because once we find a single factor, we already know the answer.

Optimizing with the Square Root

Checking every number up to n - 1 works, but it wastes effort. If n has a factor larger than its square root, it must also have a partner factor smaller than the square root. That means we only ever need to test divisors up to sqrt(n).

#include <iostream>
#include <cmath>

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i <= std::sqrt(n); i++) {
        if (n % i == 0) return false;
    }
    return true;
}

int main() {
    std::cout << std::boolalpha;  // print true/false instead of 1/0
    std::cout << isPrime(29) << "\n";  // true
    std::cout << isPrime(30) << "\n";  // false
    return 0;
}

This is dramatically faster for large numbers. Checking 1,000,003 now takes about 1,000 steps instead of a million. This is efficient because the work grows with the square root of the input rather than the input itself.

Printing All Primes Up to N

A common follow-up exercise is printing every prime in a range. We simply call our isPrime helper inside a loop:

#include <iostream>
#include <cmath>

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i <= std::sqrt(n); i++) {
        if (n % i == 0) return false;
    }
    return true;
}

int main() {
    int limit = 30;
    std::cout << "Primes up to " << limit << ": ";
    for (int n = 2; n <= limit; n++) {
        if (isPrime(n)) std::cout << n << " ";
    }
    std::cout << "\n";
    return 0;
}

Output: Primes up to 30: 2 3 5 7 11 13 17 19 23 29

Pulling the check into its own isPrime function keeps main readable and lets you reuse the logic anywhere.

Common Mistakes

Forgetting the n <= 1 check is the classic bug — without it, 1 is wrongly reported as prime. Another mistake is writing i < std::sqrt(n) instead of i <= std::sqrt(n), which misses perfect squares like 25 (25 = 5 × 5). Finally, leaving out break in the basic version still gives the right answer but does unnecessary work.

Related Articles

• C++ Loops Tutorial — for, while, and do-while explained
• C++ Conditionals Tutorial — if, else, and switch
• C++ Functions Tutorial — writing reusable helpers like isPrime
• C++ Math Functions — sqrt, pow, and the cmath library
• C++ cin User Input — reading numbers from the keyboard

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