How can I display only prime numbers in this code? - c++

Question

I'm trying to get all prime numbers in the range of 2 and the entered value using this c++ code :
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) {
result = i % b;
if (result == 0) {
result = b;
break;
}
}
cout << result<< endl <<;
}
}
the problem is that I think am getting close to the logic, but those threes and twos keep showing up between the prime numbers. What am I doing wrong?

Answer 1

I've fixed your code and added comments where I did the changes
The key here is to understand that you need to check all the numbers smaller then "i" if one of them dividing "i", if so mark the number as not prime and break (the break is only optimization)
Then print only those who passed the "test" (originally you printed everything)
#include <iostream>
using namespace std;
#include<iostream>
using namespace std;
int main()
{
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool isPrime = true; // Assume the number is prime
for (int b = 2; b < i; b++) { // Run only till "i-1" not "num"
result = i % b;
if (result == 0) {
isPrime = false; // if found some dividor, number nut prime
break;
}
}
if (isPrime) // print only primes
cout << i << endl;
}
}

Answer 2

Many answers have been given which explains how to do it. None have answered the question:
What am I doing wrong?
So I'll give that a try.
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) { // wrong: use b < i instead of b <= num
result = i % b;
if (result == 0) {
result = b; // wrong: why assign result the value of b?
// just remove this line
break;
}
}
cout << result<< endl <<; // wrong: you need a if-condtion before you print
// if (result != 0) cout << i << endl;
}
}

Answer 3

You have multiple errors in your code.
Simplest algorithm (not the most optimal though) is for checking whether N is prim is just to check whether it doesn't have any dividers in range [2; N-1].
Here is working version:
int main() {
int num = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool bIsPrime = true;
for (int b = 2; bIsPrime && b < i; b++) {
if (i % b == 0) {
bIsPrime = false;
}
}
if (bIsPrime) {
cout << i << endl;
}
}
}

Answer 4

I would suggest pulling out the logic of determining whether a number is a prime to a separate function, call the function from main and then create output accordingly.
// Declare the function
bool is_prime(int num);
Then, simplify the for loop to:
for (int i = 2; i <= num; i++) {
if ( is_prime(i) )
{
cout << i << " is a prime.\n";
}
}
And then implement is_prime:
bool is_prime(int num)
{
// If the number is even, return true if the number is 2 else false.
if ( num % 2 == 0 )
{
return (num == 2);
}
int stopAt = (int)sqrt(num);
// Start the number to divide by with 3 and increment it by 2.
for (int b = 3; b <= stopAt; b += 2)
{
// If the given number is divisible by b, it is not a prime
if ( num % b == 0 )
{
return false;
}
}
// The given number is not divisible by any of the numbers up to
// sqrt(num). It is a prime
return true;
}

Answer 5

I can pretty much guess its academic task :)
So here the think for prime numbers there are many methods to "get primes bf number" some are better some worse.
Erosthenes Sieve - is one of them, its pretty simple concept, but quite a bit more efficient in case of big numbers (like few milions), since OopsUser version is correct you can try and see for yourself what version is better
void main() {
int upperBound;
cin >> upperBound;
int upperBoundSquareRoot = (int)sqrt((double)upperBound);
bool *isComposite = new bool[upperBound + 1]; // create table
memset(isComposite, 0, sizeof(bool) * (upperBound + 1)); // set all to 0
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) { // if not prime
cout << m << " ";
for (int k = m * m; k <= upperBound; k += m) // set all multiplies
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++) // print results
if (!isComposite[m])
cout << m << " ";
delete [] isComposite; // clean table
}
Small note, tho i took simple implementation code for Sive from here (writing this note so its not illegal, truth be told wanted to show its easy to find)