Digits after decimal place.

[Winslow1342]

Hello so I got this code here to count the number of digits in a number such as 52 has 2 digits or 793 has 3 digits. I just dont understand how to do this with decimaled numbers after the decimal place... such as finding the number of digits for 72.64, It would be 4 digits but my code only says 2. Please Advise.

  #include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;


int count_digits(int num)
{
    int count = 0;
    while (num != 0) 
    {
        num = num / 10;
        count++;
    }
    
    return count;
}
int main()
{
    float f = 1;
    while (f)
    {

        cout << "Enter a Number: ";
        cin >> f;
        int i = float(f);

        cout << "Number of digits in: " << count_digits(i) << endl;
    }




    return 0;
}

[Handy Andy]

Hello Winslow1342,

Thank you for using code tags.

Be sure to read the comments.

#include <iomanip>
#include <cmath>

using namespace std;  // <--- Best not to use.


int count_digits(int num)  // <--- Defining as and "int" you loose the decimal part.
{
    int count = 0;

    while (num != 0)
    {
        num = num / 10;

        count++;
    }

    return count;
}

int main()
{
    float f = 1;

    while (f)
    {

        cout << "Enter a Number: ";
        cin >> f;

        int i = float(f);  // <--- This line unnecessary. "f" is already defined as a float and does not need to be type cast. Stuffing a "float" into an "int" loosed the decimal part.

        cout << "Number of digits in: " << count_digits(i) << endl;  // <--- You can just send "f" to the function.
    }

    return 0;
}

"<cmath>" is not required for this code. Have a look at http://www.cplusplus.com/reference/cmath/ for what is available in "cmath".

I would suggest using "double" over "float" unless your decimal numbers are small or you have a storage space problem and need the small variable.

With "f" being a floating point variable you may have a problem with the while condition. The condition, in the end, will convert what is evaluated to bool type meaning the (0) is false and (1) is true. A floating point number may not always be (0.0) or (1.0).

For your function. Define the parameter as a "double" then inside the function convert the number to a string. using "string.find" get the position of the "." then you can use simple math to figure the amount of digits to the left and right of the ".".

Just a thought and I will have to test it later.

Andy

[lastchance]

#include <iostream>
#include <string>
#include <cctype>
#include <algorithm>
using namespace std;

int main()
{
   string f;
   cout << "Enter a Number: ";   cin >> f;
   cout << "Your number has " << count_if( f.begin(), f.end(), ::isdigit ) << " digits\n";
}

I wouldn't bother counting post-decimal-point digits in float or double representation - your number only has finite floating-point accuracy. Try

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
   float f = 72.64;
   for ( int i = 2; i <= 15; i++ ) cout << fixed << setprecision( i ) << f << '\n';
}

72.64
72.640
72.6400
72.64000
72.639999
72.6399994
72.63999939
72.639999390
72.6399993896
72.63999938965
72.639999389648
72.6399993896484
72.63999938964844
72.639999389648438

[seeplus]

dont understand how to do this with decimaled numbers after the decimal place

If it's float/double, the number is really what ever number you want - see lastchance's post above.

[Winslow1342]

Handy Andy, Lastchance, and Seeplus; thank you for all your help. Got it up and running now.