Counting digits using while loop

Question

I was recently making a program which needed to check the number of digits in a number inputted by the user. As a result I made the following code:

int x;            


        

Answer 1

If x is an integer, and by "built in function" you aren't excluding logarithms, then you could do

double doub_x=double(x);
double digits=log(abs(doub_x))/log(10.0);
int digits= int(num_digits);

Answer 2

int count = (x == 0) ? 1 : (int)(std::log10(std::abs((double)(x)))))) + 1;

Answer 3

In your particular example you could read the number as a string and count the number of characters.

But for the general case, you can do it your way or you can use a base-10 logarithm.

Here is the logarithm example:

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    double n;
    cout << "Enter a number: ";
    cin >> n;

    cout << "Log 10 is " << log10(n) << endl;
    cout << "Digits are " << ceil(log10(fabs(n)+1)) << endl;
    return 0;
}

Answer 4

Bar the suggestions of reading the number as a string, your current method of counting the number of significant decimal digits is fine. You could make it shorter, but this could arguably be less clear (extra set of parenthesis added to keep gcc from issuing warnings):

while((x = x/10))
  count++;

Answer 5

You could read the user input as a string, and then count the characters? (After sanitising and trimming, etc.)

Alternatively, you could get a library to do the hard work for you; convert the value back to a string, and then count the characters:

cin >> x;
stringstream ss;
ss << x;
int len = ss.str().length();

Answer 6

Given a very pipelined cpu with conditional moves, this example may be quicker:

if (x > 100000000) { x /= 100000000; count += 8; }
if (x > 10000) { x /= 10000; count += 4; }
if (x > 100) { x /= 100; count += 2; }
if (x > 10) { x /= 10; count += 1; }

as it is fully unrolled. A good compiler may also unroll the while loop to a maximum of 10 iterations though.