Thank you for the help on is there a better way to write this

[Brad Foster]

Never mind I went over the whole code and found my mistake, I did not use
int sumSquareOdd=0, I either did not finish it or I erased it while I was trying something else. I knew it was not keeping count correctly

now I have a couple of questions
on JonLT’s code
1. int sumSquareOdd = 0;
2.
3. for (int n = firstNum; n <= secondNum; n++)
4. {
5. sumSquareOdd += n * n * (n%2); // if n is even, 0 will be added (unchanged)
6. }
how did it know that I wanted odd numbers without (n%2==1)?

On the code sumSquareOdd += n * n * (n%2); why is += used instead of just =?

[JonLT]

Never mind I went over the whole code and found my mistake, I did not use
int sumSquareOdd=0, I either did not finish it or I erased it while I was trying something else. I knew it was not keeping count correctly

now I have a couple of questions
on JonLT’s code
1. int sumSquareOdd = 0;
2.
3. for (int n = firstNum; n <= secondNum; n++)
4. {
5. sumSquareOdd += n * n * (n%2); // if n is even, 0 will be added (unchanged)
6. }
how did it know that I wanted odd numbers without (n%2==1)?

On the code sumSquareOdd += n * n * (n%2); why is += used instead of just =?

if n is even the statement (n%2) will be 0 (false)

foo += bar is the same as foo = foo + bar

[tburger]

As John illustrated above, the "+=" syntax will add the value that is calculated on the right side of the assignment to the existing value of the variable on the left.

This is often helpful in keeping a running total (i.e. keeping a total sales figure as you run through a large list of numbers).

For future reference, you may also use "*=" and "-=" in your equations.

Tom

[Brad Foster]

I don't know how the program knew that I wanted only odd #s without (n%2==1).

see in all my other code I had to use (n%2==0) or (n%2==1)

for(n=firstNum;n<=secondNum;n++)
{
if(n%2==1)
{
cout << "The odd numbers are: " << n << endl;
}

//since I want the sum of the even number I don't need the else statement
}
for(n=firstNum;n<=secondNum;n++)
{
if(n%2==0)
sum += n;
}
cout<< "The sum of all even numbers between your two numbers: " << sum << endl;

in the above two examples I needed those statements two tell the statements to use the odd or even...........I am trying to figure out if it is taking it from one of these two statements.

[tburger]

As you know, the statement (n%2) can only produce two values: 1 if odd or 0 if even.

Therefore, the statement can essentially function as a true/false trigger in your program.

In both of the examples above, you could have just used (n%2) and it would have worked.

Let's break it down:

If the number n is 5, then (n%2) becomes (5%2). This will return a value of 1.

The statement

if(n%2)

is now seen as

if(1) .

Therefore, the program continues with the following code.

If the number were even, however, the program would see

if(0) and ignore the next "if" statments.

Tom

[weaknessforcats]

Everyone please use code tags.