Checking if value is float or int

[Nisheeth]

I want to create a program that will find the factors of a given number.
I have thought up of this concept:
The User will enter a value (say x).
The program will divide the number by 2 then 3, than 4 ... (until x/2)
The program will then publish all the values that were whole numbers.

Now I wanted help with two things:
1. How do I make the program see that whether the value is int or float?
2. How do I make the program publish the values that returned the int value?

Thanks in advance

[Zeillinger]

I believe that in your case, checking the following would be enough: if (num == floor(num)) (using math.h/cmath).

[Nisheeth]

Could you please elaborate?
what does floor (num) mean?

[Zeillinger]

It rounds down the number, if you aren't familiar with the standard math library I would recommend visiting http://cplusplus.com/reference/clibrary/cmath but looking at your second topic, I noticed you don't actually need that. As hamsterman correctly mentioned, you should use the modulus operator (%).

[Nisheeth]

O.K. I got how to verify that whether the value is a factor or not.

Now how do I publish those values after the computation is over?

[Zeillinger]

By "publish", do you mean print the values? You'll just need to add basic output to your if statement.

[Nisheeth]

By "publish", do you mean print the values? You'll just need to add basic output to your if statement.

Yes. But say there are 5 values that give the result as 0. I don't know which values they are. But I want the program to be able to print those values in the output. Now How do I do that?

[Zeillinger]

That's really hard to follow that way. I don't know what variables or conditions you use so the best way to proceed would be posting your code here.

[Nisheeth]

#include <iostream>
#include <string>

using namespace std;
//Variables
	int number;	//The number whose factors are to be calculated
	int divider = 2;     //The Divider
	int factor;	//Potential Factors
	int factors;	//Actual Factors

	//Exit Variables
	char x = 'N';
	char h = 'N';

int main ()
{
	do 
	{
	//Asking for input:
	cout << "Please enter the value whose factors you want to calculate:" << endl;
	cin >> number;
	cout << endl;
	
	//Calculatin the value
	do
	{
		factor = number % divider;
		//Divider condition
		do 
		{
		if (divider < number/2)
		{
			divider = divider + 1;
		}
		}while (divider < number/2);

	} 
	while (divider > number/2);
	
	//Giving the factors:
	cout << "The factors of " << number <<" are:" << endl;
	cout <<"1" << endl;
	cout << number << endl;

	//Code for exiting the program.. irrelevant to the its functioning...

	//Terminating the Program
	return 0;
}

In giving the factors part, I have put up 1 & the number as the factors. I don't know how to do so for other factors.

[Zeillinger]

Could you give an example of input + output?

[Nisheeth]

O.K.
Lets say I input

50

Then I want the output to be:

The Factors are:
1
2
5
10
25
50

[Zeillinger]

You over-complicated it. You should simply loop through the numbers 1...n/2 and print the once that n can be divided by without a remainder. Those will be the factors.

// ... receive `n` from the user, etc.
cout << "The factors of " << n << "are:" << endl;
for (int divider = 1; divder <= n/2; divider++) {
	if (n % divider == 0)
		cout << divider << endl;
}
cout << n << endl;

[Nisheeth]

Thanks. Can you explain what you did here for future reference?

[Zeillinger]

As I said, the loop goes from 1 to n/2, increasing `divider` and for every value of `divider` it is to check if n can be divided by `divider` using the modulus operator. if it can, `divider` is indeed a factor of n so we print it. Also, n can always be divided by n so we print it as well (just like you did).

Hopefully it's clear now :)

[Nisheeth]

Yes. Its perfectly clear now. Thanks a lot

[Duthomhas]

for (int divider = 1; divider <= n/2; divider++) {

[Zeillinger]

Duoas, you're right. I've missed that...