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;    
cout << "Enter a number: ";
cin >> x;
x /= 10;
while(x > 0)
{
  count++;
  x = x/10;
}

From what I can tell (even with my limited experience) is that it seems crude and rather unelegant.

Does anyone have an idea on how to improve this code (while not using an inbuilt c++ function)?

I am assuming `x` is an `int` or else this will *never* reach `0` — Naftali, Jul 06 '11 at 19:08
"Crude and unelegant" is what you're stuck with if you can't use inbuilt functions. — Mark Ransom, Jul 06 '11 at 19:17
What do you mean, no inbuilt functions? On ARM CPUs for example, the division operator is an inbuilt function. — Zan Lynx, Jul 06 '11 at 19:22
Sorry, gotta clarify. I am assuming there is a method to simply count the number of digits in an integer using some function in the C++ standard library(i may of course be wrong). To learn more about C++ I have decided to try to make most of the code myself, just to get a feel of how stuff works hence the attempt at try to count the number of digits "manually". — E.O., Jul 06 '11 at 19:27

Zan Lynx · Accepted Answer · 2011-07-06T19:26:50.380

12

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;
}

edited Jul 06 '11 at 19:26

answered Jul 06 '11 at 19:08

Zan Lynx

53,022
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3

`log10(10) == 1`, so I think you need `floor(log10(n)) + 1`. – Oliver Charlesworth Jul 06 '11 at 19:17
3

Try x=0, or x=-1, or x=999.999. – David Hammen Jul 06 '11 at 19:26
1

@David: Let us define this program's domain as integral numbers from 1 to infinity. Although I did fix it to handle negative numbers. :-) – Zan Lynx Jul 06 '11 at 19:32
@David: Well, leading zeros disappear from a number. Generalizing from this we can claim that all leading zeros disappear. Therefore, 0 is equal to nothing at all and is properly known as 0 digits. :-) – Zan Lynx Jul 06 '11 at 20:12

David Hammen · Answer 2 · 2011-07-06T19:22:13.120

5

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

edited Jul 06 '11 at 19:22

answered Jul 06 '11 at 19:14

David Hammen

32,454
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No, this gives the right result **except** for x = 0, corrected. – David Hammen Jul 06 '11 at 19:21

score 2 · Answer 3 · answered Jul 06 '11 at 19:08

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();

score 1 · Answer 4 · answered Jul 06 '11 at 19:14

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);

answered Jul 06 '11 at 19:14

Dan

12,157
12
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score 1 · Answer 5 · answered Jul 06 '11 at 20:14

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.

score 0 · Answer 6 · answered Oct 09 '13 at 14:01

0

#include<iostream>
using namespace std;
int main()
{
int count=0;
    double x;
    cout << "Enter a number: ";
    cin >> x;
    x /= 10;
    while(x > 1)
    {
      count++;
      x = x/10;
    }
    cout<<count+1;
}

answered Oct 09 '13 at 14:01

reza

1

score -1 · Answer 7 · answered Jul 06 '11 at 19:13

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++;

Counting digits using while loop

7 Answers7

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