Trying to make a program to return only prime numbers and I'm lost at else if (x % x == 0)

[kind garden]

`#include <iostream>

using namespace std;

int main() {

//Variables

int min_value, max_value;

//Ask user for min and max value and check

cout << "Please enter a minimum value bigger than 1: " << endl;
cin >> min_value;

while (min_value <= 1){
    cout << "Please enter a minimum value bigger than 1: ";
    cin >> min_value;
}

cout << "Please enter a maximum value: " << endl;
cin >> max_value;

while (min_value == max_value || min_value > max_value){
    cout << "Please enter a valid maximum value: ";
    cin >> max_value;
}

//Out for loop

for (int x : {min_value, max_value}){ //Declare range

    if (x == 2 ) {
        cout << x;
        x++;
    }
    else if (x == 3){
        cout << x;
        x++;
    }
    else if (x % x == 0){
        cout << 6*(x) - 1 << endl;
        x++;
    }
}

return 0;

}`

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[kind garden]

Theres a 2 formulas to find prime numbers and its 6x-1 and 6x+1

Issue is, when I inserted a min value of 2 and a max value of 15 the output should be: 2,3,5,7,11,13. But it only outputs 2 and for some reason 89 and only those two numbers.

So now my question is, how do I first start the counter to increase by an increment of one, and second make it only stay in the range of the specified max value

[final eagle]

Those formulas are not sufficient for finding primes. All primes greater than 3 can be expressed with one of those two formulas, but it also works with some non-primes. Consider for example when x = 6. That yields 6 * 6 - 1 = 35, which is not prime.

It's still possible to use those formulas to quickly eliminate some candidates for prime, but those by themselves are not sufficient for finding primes.

Your code is also quite bizarre, but I'm not sure if it's worth the effort to try to fix it when the underlying algorithm won't work.

[kind garden]

What should I do instead then? I only have knowledge about for loops, do while loops, while loops, and if statements.

[final eagle]

It's not the most efficient solution, but you can use a for loop to iterate through all the values from min_value to max_value and check if they are divisible by any other number less than them. That could be accomplished using nested for loops.

There are far more efficient ways to do this, but you do not seem ready to attempt optimizations.

[kind garden]

Okay. I’ll give it a try when I can. Thank you.

Was the first for loop alright though

The for (int x : {min,max})

[final eagle]

No. That is not how a normal for loop is written. The code does run, but it iterates only twice. Once for the min_value and a second time for the max_value. The normal code would be:

for(int x=min_value; x<=max_value; x++){
  //loop code body    
}

Also note that it is not normal to modify the iterator inside the loop. This is legal, but it makes reasoning about the code much harder and can cause many bugs that are difficult to understand without using a debugger.

[azure badge]

@kind garden Has your question been resolved? If so, run !solved :)

[kind garden]

Ohh okay I see. Thank you.

[kind garden]

for (int x = min_value; x <= max_value; x++){ for (int num = 2; num <= max_value; num++) if (x % num == 0) cout << x << endl;

@final eagle this is what i have for the function currently but I don't know what I'm doing wrong as it still outputs 2-15 but multiple of them

i believe im going in the right path but have the if statement wrong

[final eagle]

The inner loop should only iterate from 2 up to but not including x. You actually could optimize that to 1/2x. Iterating up the max_value on the inner loop is not correct. Also, instead of outputting every number that is divisible, you should be outputting the numbers that aren't.

It's technically possible for the inner loop to iterate up to max_value, but it's not only inefficient. You will need to exclude the comparison when x is equal to num, because a number is always divisible by itself. It would be simpler to not iterate as high as x.

[worldly urchin]

https://en.m.wikipedia.org/wiki/Sieve_of_Eratosthenes

take a look at this fast method of determining primes

this is what i would use

you can store the primes in a std::vector and index them when you need them

if there are not enough primes in std::vector, then generate more

[kind garden]

What if statement would I say in order to output the ones that aren’t?

[final eagle]

Start with the assumption that the number is prime. If the number is determined to be divisible inside the loop, change it to being non-prime. At the end of the loop, if the original assumption of it being prime was not proven wrong, then output the number.

[final eagle]

Did you look at the code that the OP wrote? They are very far from being ready to do any kind of optimizations.

[kind garden]

Mb didn’t see this, wifi was bad at the place I was at

I may not be able to do that

[kind garden]

Ill try that when I get home

[kind garden]

for (int x = min_value; x <= max_value; x++){ for (int num = 2; num <= x; num++) if (x % num == 0) cout << "" << endl; else if (x % num == 0) cout << x; }

what am I doing wrong?

I've tried a lot of combinations and I either get no ouput at all or get every number repeated like 10 times

[final eagle]

You are deciding too soon. You must either go through all iterations of the inner loop without finding a divisible number or you must at least iterate until you find one. You are instead making a decision on each iteration.

Also, your if and else if conditions are identical.

[kind garden]

So if I only do one if (x % num == 0) and then that cout

and then another if statememt?

[kind garden]

yeah ik, i was just trying anything and everything

[final eagle]

That would still be deciding too soon.

[kind garden]

I'm not really following what you mean

Are you saying don't cout?

Since its too early of a decision

[final eagle]

I was very precise in my words. You must not decide if a number is prime on each iteration, but you can decide it's non-prime on any iteration.

The decision of prime can only happen after all iterations of the inner loop.

[kind garden]

So if I did something

that

went through all the numbers

and checked if it was and wasn't

if it wasn't, it would append it to some sort of list

and then cout it

sorry messed up wording

[final eagle]

Why did you append them to a list? This does not help in any way, unless you are making a list of verified primes as an optimization, but you are not ready to optimize.

[kind garden]

I don't know

[final eagle]

You are guessing instead of trying to think through the logic.

This exercise is actually a logic puzzle.

[kind garden]

Hm okay

So rn I should focus on iterating through the loop and decide if its not prime

and see from there

[final eagle]

Yes.

[kind garden]

Okay

Also is there a reason

why when I originally had an output

it would output a bizzare amount of numbers

instead of just a singular number

for example

if i inputted 2 - 15

it would output 2 - 15 but like 20-30 times

[final eagle]

You outputted something on every iteration of every loop instead of waiting until you found a prime.

[kind garden]

okay

thank you

[azure badge]

@kind garden Has your question been resolved? If so, run !solved :)

[kind garden]

I finally understood what this REALLY meant

and did this

`//Out for loop
for (int x = min_value; x <= max_value; x++){

    //Assume that it is a prime number from the get go and try to falisfy it

    bool prime = true;

    for (int num = 2; num <= x; num++)

        if (x % num == 0){
            prime = false;
            break;
        }

    if (prime)
        cout << x << " ";
}`

i think this is the correct answer but i dont know why it still outputs nothing

my teacher is saying it has something to do with the square root of soemthing in the instructions

but im not too sure what that means at all

[final eagle]

There is only one minor problem. The inner loop must not iterate up to x. It can iterate to 1 less than x, but comparing x to itself will always result in it being divisible. I mentioned this yesterday.

The usual solution iterates to 1/2 x.

[kind garden]

Okay okay

I’ll try it when I’m done eating

[kind garden]

It works

Thanks for all the help man

!solved