How to only show decimal places and not int?

Hey I'm really stuck on how to show only decimal places in a number.

For example, if I had 1.234 I would only want to output .234

Hmm... one way that I can think of is to #include <cmath> then say, call std::fmod(1.234, 1.0)

Yeah... laserlight is correct....

And The another way is to subtract the floor of the number from the original number..

I mean, 2.345 - 2 = .345
say a =2.345
a - floor(a);

floor(); is available in math.h header file...

So no worries..:-)

Hmm well basically the user inputs Rmin and Rmax values. Then the mean is computed.

I have three variables, A B and C.

A B & C are parts of the mean in scientific notation. For example,

7.4 x 10^1 (7=A, 4=B, 1=C)



My code is:

Code:

---

#include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
    int A, B, C, D;
    double x, y, mean;
       
                //Entering Values
                cout << "Enter RMin: ";
        cin >> x;
        cout << "Enter RMax: ";
        cin >> y;
        mean=(x+y)/2;
       


                //Print out mean
                cout << "Mean: " << mean << endl;


                //B
                ????????????


                //C
                C=log10(mean);
 
                //A
        while (mean>10)
        {
        A=mean/=10;
        }


                //Printing A B C D
        cout << "A: " << A << endl;
                cout << "B: " << B << endl;
                cout << "C: " << C << endl;


        return 0;
}

---

The ???? represents where I'm stuck.

I also tried using a loop

Code:

---

while (mean>10)
{
B=mean%10;
}

---

Edit: Ignore D (Nothing to do with my problems)

Code:

---

#include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
    int A, B, C, D;
    double x, y, mean;
       
                //Entering Values
                cout << "Enter RMin: ";
        cin >> x;
        cout << "Enter RMax: ";
        cin >> y;
        mean=(x+y)/2;
   


                //Print out mean
                cout << "Mean: " << mean << endl;




                //C
                C=log10(mean);




                //D
                D=(100*(y-mean))/mean;




                //A
        while (mean>10)
        {
        A=mean/=10;
        }


                //B
                B=(mean - floor(mean))*10;


 


                //Printing A B C D
        cout << "A: " << A << endl;
                cout << "B: " << B << endl;
                cout << "C: " << C << endl;
                cout << "D: " << D << endl;


        return 0;
}

---

It works for smaller numbers: Examples:

Rmin=60
Rmax=80
Mean=70
A=7
B=0
C=1
D=14


But for bigger numbers:
Rmin=51000
Rmax=61000
Mean=56000
A=5
B=5
C=4
D=8

The B value for bigger mean's is always wrong...

if you literally only want to show the decimal part then you may as well just convert it to a string and trim off whatever precedes the decimal point- if you need it for something mathematical then your problems remain

B needs to be an int value of whatever is after the decimal.

So for 7.4:

A = 7
B= 4 (not 0.4)

Sorry, I'm lost as to what you're asking about now. What are you actually trying to do?

Basically, the user inputs two values which are added and divided by 2 to get the mean.

Using the mean, the values of A B and C are obtained. For example,

184 = 1.84 x 10^2 (in scientific notation) - A = 1, B=84 C=2
74 = 7.4x10^1 - A=7, B=4, C=1

I'm not sure how to code B.

I think you may need to use fixed point arithmetic instead of floating point arithmetic, otherwise it may be almost impossible to correctly determine the value of B.

It may be a good idea to tell us the exact requirements and why you are doing this in the first place. I'm thinking of things like: what is the range of input? Can you read the input as a string or an int instead of a double? Would formatting the input into a string and extracting the parts be okay, as rogster001 suggested?

in the examples from post #10 you could just use the left hand value to extract A and B at least then you dont have to consider the decimal point - But then we have no idea what your overall aim is.