code to multiply decmils only

[nu123]

hi there I'm trying to write a program that will only multiply the decimal in the number i enter by 9. so for example if i enter 2.345 the program will only multiply .345 by 9 and ignore the 2, i have a basic program for multiplication but i can not figure out how to modify it to get it to do this

#include <iostream>
using namespace std;

int main() 
{
    double firstNumber, answer;
    cout << "Enter a number above 0: ";

    // Stores two floating point numbers in variable firstNumber and secondNumber respectively
    cin >> firstNumber;
 
    // Performs multiplication and stores the result in variable productOfTwoNumbers
    answer = firstNumber * 9;  

    cout << "Product = " << answer;    
    
    return 0;
}

[rghrist23]

You would need to use std::modf. This will break the double into two doubles where one contains the whole integer and the other has the decimal part.

[nu123]

where about would I insert it into the code. i have tried in a few places but i keep getting errors. I'm fairly new to c++ so don't have much of a grasp on it yet. thanks for the help

[wildblue]

There's an example of modf being used here:

http://www.cplusplus.com/reference/cmath/modf/

The <cmath> library also has floor (and ceil) functions to round down (or up) a number to the nearest integer. If you subtracted the floor of the original number from itself, you'd be left with the fractional part.

[nu123]

that's great thanks, I'm after finding a code and have it doing close to what i want, so now I'm just trying to get the program to multiply the decimal point by 9 and ignore the rest.
code I'm after finding is
#include <iostream>
#include <iomanip>
int main()
{
printf("enter number");
float f=1;
while (f)
{
std::cin >> f;
int i = (int)f;
std::cout << i << '\t' << std::setprecision(3) << f - i << std::endl;
}
return 0;
}

[texasbeef]

I like learning how to do things without library functions. It helps me understand what is going on in the code. This is probably a noob answer but I can do the calculation you are looking for like this:

#include <iostream>

using namespace std;

int main()
{
   double num1 = 2.345;
   int num2 = (int)num1;
   double num3 = (num1 - (double)num2);
   cout << num3 << " * 9 = " << num3 * 9 << endl; 
   
   return 0;
}

output:

0.345 * 9 = 3.105

[Chibiken]

output:

0.345 * 9 = 3.105

What is wrong with that result?

[rghrist23]

I do not think anything is wrong with @texasbeef results. He was showing a way that does not use cmath floor, but achieves the similar method and same result.