Home work help

[jackelinblack]

I'd like you guys to look over my code and see if I did anything wrong. I missed my class, so i'm just not sure if I answered the question correctly. I personally think it's totally wrong but idk.

question number 1.
Write a program that reads a set of integers and then finds and prints the sum of the even and odd integers.

  #include <iostream>
using namespace std;

const int num = 10;
int main() {
    
    int n = 1;
    
    while (n <= num) {
        if (n % 2 == 0) {
            int sume;
            sume = sume + n;
            n++;
            cout << "The value of your even numbers is : " << sume << endl;
        }
    
        else{
            int sum;
        sum = sum + n;
            n++;
            cout << "The value of your odd number is : "<< sum << endl;
        }
    }
    
    return 0;

[anup30]

int n, even , odd;
cout << "enter number: ";
while(cin >>n)
{	
if( n == n/2*2 )
	even+=n;
else	
	odd+=n;
cout << "enter another number: ";	
}
cout << "\nsum even = " << even << "\nsum odd =" << odd << endl;	

[jackelinblack]

How does your formula (n== n/2*2) work to find an even number?

if n =2 then 2==2/4 which is 2==1/2?

I haven't learned the even+=n part, could you write the long version of that formula? is it even = even + n?

[MiiNiPaa]

How does your formula (n== n/2*2) work to find an even number?

if n =2 then 2==2/4 which is 2==1/2?

Arrr. ORDER OF OPERATIONS! Multiplication and division has the same priority and executes left to right.
2/2*2 → 1*2 → 2.

However using modulo operator is faster and easier n % 2 == 0 → even

I haven't learned the even+=n part, could you write the long version of that formula? is it even = even + n?

Yes

[deathslice]

actually jackelinblack it would 2==2 because 2/2 = 1 multiply by 2 and you get 2. Also even += n is the same thing as even = even +n so you're correct. Remember that order of operation is important.

[jackelinblack]

oh ok thanks. In general how do you know which way is faster?

[MiiNiPaa]

Two trivial operations against one. It could possibly be optimised by compiler, but it is not guaranteed. Additionally n==n/2*2 is cryptic and requires consulting an n type to determine what it actually does.

[deathslice]

from what I can remember, if you have a finite list and you want to reiterate that list over a loop, I would say that using modulus operator is more suitable for that situation than doing x+= y. Sometimes it's not about which is fastest but what is more suitable, convenient and efficient I.E expressing the remainder other than a decimal, having a list and wanting to see how many are left over etc etc. Did I answered your question or no? If not, then I apologize for wasting your time.

[jackelinblack]

lol no but it's ok.