Count digits of number without loop C++

Question

I have to count the number of digits in a number.

I divide the number to 10 until I get 0. Each iteration increments the counter.

int num;
cin>>num;  
while(num > 0)  
{  
  counter++;
  num = num / 10;   
}

The challenge is not using any loops or recursion, just an if statement.

Is it possible?

@dolan, Nice idea! Like! but it is also forbidden in this case :-( — yael aviv, Sep 08 '14 at 09:35
@yaelaviv `log10(n)+1` where `n` is the number. For example, `log10(1234) + 1` is 4. — Maroun, Sep 08 '14 at 09:36
If you can't use recursion you should state that in the question. — juanchopanza, Sep 08 '14 at 09:36
@yaelaviv Because of math :D `log10` of thousands is 3.something, `int` value is 3, hence you need the +1. `log10` of hundreds is 2 and so on.. — Maroun, Sep 08 '14 at 09:39
Actually, I _thought_ I'd seen something similar before: see http://stackoverflow.com/questions/1068849/how-do-i-determine-the-number-of-digits-of-an-integer-in-c/1068937#1068937 — paxdiablo, Sep 08 '14 at 09:56

MoonBun · Accepted Answer · 2014-09-08T11:23:00.933

13

counter = log(num) / log(10)

in c++:

#include <cmath>
....
counter = num == 0 ? 1 : log10(std::abs(num)) + 1;

what you want is the log function.

cplusplus - log10

cplusplus - std::abs

edited Sep 08 '14 at 11:23

answered Sep 08 '14 at 09:35

MoonBun

4,322
3
37
69

1

Still broken for `num == 0` though. – Paul R Sep 08 '14 at 09:40
@Vladp, Does log10 have to use double variables? – yael aviv Sep 08 '14 at 09:52
2

This doesn't work for num == int min (on a 2s complement machine). I'm not sure what standard says std::abs does in that case but the answer it gives is clearly wrong. [see http://ideone.com/exvOQK] – Mike Vine Sep 08 '14 at 09:56
@Mike, `std::abs()` inherits from the C `stdlib` header, which specifies undefined behaviour if the result cannot be represented. – paxdiablo Sep 08 '14 at 11:52

score 3 · Answer 2 · answered Sep 08 '14 at 09:36

3

Easy way although somewhat expensive, turn your number to string and take its size like the example below:

#include <iostream>
#include <string>

int main() {
  int i = 1232323223;
  std::string str = std::to_string(std::abs(i));
  std::cout << "Number of Digits: " << str.size() <<std::endl;
}

LIVE DEMO

answered Sep 08 '14 at 09:36

101010

41,839
11
94
168

Doesn't work for int min (on a 2s complement machine) – Mike Vine Sep 08 '14 at 09:51

Paul R · Answer 3 · 2014-09-08T10:08:46.593

3

One way is to use sprintf, since it returns the number of characters emitted:

int digits(int n)
{
    char s[32];
    int len = sprintf(s, "%d", n);
    if (n < 0) len--; // NB: handle negative case
    return len;
}

edited Sep 08 '14 at 10:08

answered Sep 08 '14 at 09:36

Paul R

208,748
37
389
560

I think `s` should be `temp` (or vice versa), yes? – paxdiablo Sep 08 '14 at 09:52

paxdiablo · Answer 4 · 2014-09-08T10:11:52.893

Sure it's possible, something like (for 32-bit numbers):

int numDigitsU (unsigned int n) {
    if (n <         10) return 1;
    if (n <        100) return 2;
    if (n <       1000) return 3;
    if (n <      10000) return 4;
    if (n <     100000) return 5;
    if (n <    1000000) return 6;
    if (n <   10000000) return 7;
    if (n <  100000000) return 8;
    if (n < 1000000000) return 9;
    /*      4294967295 is 2^32-1 - add more ifs as needed
       and adjust this final return as well. */
    return 10;
}

The signed variant is a little trickier since the sign is reversed first, and you have to watch out for MININT:

int numDigitsS (int n) {
    if (n == MININT) n = MAXINT;  // same number of digits, usually.
    if (n < 0) n = -n;            // reverse sign.
    return numDigitsU (n);        // call the unsigned variant.
}

Just adjust the highest comparison value and return value based on the size of your largest unsigned int.

This should work with all allowed negative codings: two's complement, ones' complement, and sign/magnitude.

Here you are actually unrolling the loop. OK is faster than using a for loop but quite verbose also. — 101010, Sep 08 '14 at 09:45

Peter · Answer 5 · 2014-09-08T09:49:57.483

3

It's actually quite easy if you consider that the maximum size of int is finite. Just check if the number is larger than 10, larger than 100, larger than 1000, etc. You could even do binary search.

int num = abs(number);
if (num < 10000)
{
    if (num < 100)
        return num < 10 ? 1:2;
    else
        return num < 1000 ? 3:4;   
}
else
{
    ...
}

edited Sep 08 '14 at 09:49

answered Sep 08 '14 at 09:42

Peter

5,608
1
24
43

`you could even do binary search` how? – yael aviv Sep 08 '14 at 09:45
@yael Edit: Provided an example – Peter Sep 08 '14 at 09:50

score 0 · Answer 6 · edited Jun 01 '20 at 14:28

0

You can find the length of an integer just by doing it as follows:

int countDigits(int *num){
    int count =0;
    while(*num>0){
        count++;
        *num /=10;
    }
    return count;
}

edited Jun 01 '20 at 14:28

David Buck

3,752
35
31
35

answered Jun 01 '20 at 13:44

crispengari

7,901
7
45
53

Count digits of number without loop C++

6 Answers6

Linked