Quote:Original post by Haptic
Well you already have a while loop there that goes through all of the numbers (albeit backwards), so for question 4, why not just add a few lines that calculates and outputs the square of the current number.

To calculate if a number is odd or not you need to divide it by 2 and get the remainder. In C, this is done with the % operator. So 3%2 = 1.
If the answer is zero, it's even, if it's 1 its odd. You just need to add this condition to your loop.

As a side note - you are setting the value of 'counter' before you set the value of secondNum which is a bug.

Quote:Original post by SilentEcho
The reason you are having trouble with setw is because it's part of iomanip, so in order for your code to compile you will need to add #include<iomanip> to the top of your file.

It basically tells the stream to output the next piece of data with a set width you provide to the setw function. It will output that many characters worth of width regardless of whether or not the data fills it all up. If it doesn't, it will output blank space to fill in the rest. You can also align the output left or right, say...

cout << setw( 5 ) << right << someInt;
// will output value in someInt to the console with width of 5, right aligned

(right and left are members of std, so you either need to use the namespace std or type the fully resolved std::right, std::left)

Quote:Original post by Haptic
% returns the 'remainder' of a division.

So for example, if you divide 11 by 2:
2 goes into 11 five whole times - with 1 remainder.
So 11%2 will return 1.

*** Source Snippet Removed ***

Some more examples:
8 % 3 = 2
4 % 1 = 0
12 % 5 = 2

    if (counter%2 == 1)    remainder = odd;    if (counter%2 == 0)    remainder = even;

cout << setw(odd) << odd << endl;

Quote:Original post by Ultimatepuff
I have no idea where to start in putting it into my code.

Quote:Output all odd numbers between firstNum and secondNum.

if (secondNum % 2 == 1) remainder = odd;else remainder = even; // we don't need to test again: if it's not odd, it has to be even.

firstNum = 10 secondNum = 21output: 11, 13, 15, 17, 19firstNum = 11secondNum = 28output: 13, 15, 17, 19, 21, 23, 25, 27firstNum = 7secondNum = 16output: 9, 11, 13, 15

if (secondNum % 2 == 1) counter = secondNum - 2; // odd;else counter = secondNum - 1; // even

while (counter >= firstNum){  cout << counter << "\n";  counter -= 2;}

Quote:Original post by WanMaster

Quote:Original post by Ultimatepuff
I have no idea where to start in putting it into my code.

Haven't followed the entire thread, but I assume this is about problem 2:

Quote:Output all odd numbers between firstNum and secondNum.

You already know how to find out whether a number is odd or even. There are more ways to do it, but the suggested manner should do fine.

We start the loop with the highest number. First, let's find out if this number is odd or even:

if (secondNum % 2 == 1) remainder = odd;else remainder = even; // we don't need to test again: if it's not odd, it has to be even.

Let's see some examples of what the output should look like in several cases:

firstNum = 10 secondNum = 21output: 11, 13, 15, 17, 19firstNum = 11secondNum = 28output: 13, 15, 17, 19, 21, 23, 25, 27firstNum = 7secondNum = 16output: 9, 11, 13, 15

As you may notice, we can start our counter at either secondNum - 1 if the number is even, or secondNum - 2 if the number is odd. Let's use that in our code:

if (secondNum % 2 == 1) counter = secondNum - 2; // odd;else counter = secondNum - 1; // even

Again, as you can see in the example outputs, the only thing left to do is to decrease this counter by 2 every iteration, until it is equal or smaller to firstNum. There's no need to keep checking for the odd/even property, because if say, a number is odd then every time you add or substract a multiple of 2, you get a new number that is also odd. Same goes for even numers. The codez:

while (counter >= firstNum){  cout << counter << "\n";  counter -= 2;}