number of digits in number?

**MKashlev** · 08-10-2002

Is there any function that calculates the number of digits in an int?

**subdene** · 08-10-2002

The best way i can think of doing this, is to convert the int to a string using itoa(), then use a while loop with a condition to break when the first '\0' is encountered. Therefore, the string will have to be null terminated before you use itoa, within the while loop you would have a counter incrementing until the condition becomes false. You then have the number of digits from the int.

**subdene** · 08-10-2002

Code:

#include <iostream.h>

int main(void)
{
  int Num, Digit=0;
  static char *StringNum;

  cout << "Please enter number: ";
  cin >> Num;

  StringNum = new char[sizeof Num + 1]; //'\0 +1'

  itoa(Num, StringNum, 10);

  while(StringNum[Digit] != '\0')
    Digit++;

  delete StringNum;
  StringNum=NULL;

  cout << "\nThe number: " << Num << ". Contains " << Digit << " digit(s).";
  getchar();

  return 0;
}

**Crimpy** · 08-10-2002

Peform a loop that increments a counter variable and divide the number or a copy of the number by the base until it reaches 0.

Code:

#include <iostream.h>

int main ( )
{
    int number = 64789, copyof;
    int count = 0;

    copyof = number;
    while (copyof){
        count++;
        copyof /= 10;
    }
    cout<<"There are "<<count<<" digits in "<<number<<".\n";

    return 0;
}

silentstrike · 08-10-2002

I got a little bored. Looks like division by 10 is the way to go.

Code:

#include <cmath>
#include <ctime>
#include <iostream>
#include <sstream>
#include <iomanip>

using namespace std;

const int test_iterations=int(1e7);

int numDigitsWithDivBy10(int n) {
	if (n == 0) return 1;
	int count=0;
	while (n) {
		count++;
		n/=10;
	}
	return count;
}

int numDigitsWithLog(int n) {
	if (n < 0) n = -n;  // log of negative illegal
	else if (n == 0) return 1; // log 0 illegal
	return int(log10(double(n)))+1;
}

int numDigitsWithSStream(int n) {
	if (n < 0) n = -n; // don't want to count negative sign in length of string
	ostringstream out;
	out << n;
	return (int) out.str().length();
}

// test for correct results
void test(int n) {
	std::cout << setw(13) <<  n << '\t' << numDigitsWithLog(n) << '\t' << numDigitsWithSStream(n)
	<< '\t' << numDigitsWithDivBy10(n) << endl;
}

int main() {
	int i;

	for (i = 1; i < INT_MAX/10; i*=10) {
		test(i);
		test(i - 1);
		test(-i);
		test(-i + 1);
	}

	time_t start=time(NULL);
	for (i = 0; i < test_iterations; ++i) {
		numDigitsWithLog(i);
	}
	cout << "numDigitsWithLog for " << test_iterations << " took " << time(NULL) - start << " seconds " <<endl;

/*	start=time(NULL);
	for (i = 0; i < test_iterations; ++i) {
		numDigitsWithSStream(i);
	}
	cout << "numDigitsWithSStream for " << test_iterations << " took " << time(NULL) - start << " seconds " <<endl;
*/

	start=time(NULL);
	for (i = 0; i < test_iterations; ++i) {
		numDigitsWithDivBy10(i);
	}
	cout << "numDigitsWithDivBy10 for " << test_iterations << " took " << time(NULL) - start << " seconds " <<endl;

	return 0;
}

**SilentStrike** · 08-10-2002

Forgot to log in, so I can't edit it

. Oh well.

**d00b** · 08-11-2002

Originally posted by silentstrike
I got a little bored.

This is what you do when you're bored?

**SilentStrike** · 08-11-2002

Yeah.. pretty enjoyable

**MKashlev** · 08-11-2002

what is sstream header in silentstrike's code?

**SilentStrike** · 08-11-2002

It for NumDigitsWithSStream. It basically lets you write to a string with <<. It similair to cout (in that it is an output stream), but instead of writing to the console, it writes to it's own private buffer, which can be converted to a string with the .str() method.

ps_th · 08-11-2002

or you can use log10().
Here's an example

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
int n=1078;
cout<<n<<" has "<<static_cast<int>(log10(n))+1<<" digits\n";
return 0;
}

**SilentStrike** · 08-11-2002

I had that in my post already.. benchmarked it, ran about 2.5 times slower than the division by 10 method.

**Salem** · 08-11-2002

Well if you're out for speed, try this.

It's a bit long compared to some of the others, but it is rather quick

Would be somewhat unpleasant to extent to 64 bits though

Code:

int num_len ( int n ) {
    if (n < 0) n = -n;
    if ( n <= 99999 ) {
        if ( n <= 999 ) {
            if ( n <= 9 )  return 1;
            if ( n <= 99 ) return 2;
            return 3;
        } else {
            if ( n <= 9999 ) return 4;
            else             return 5;
        }
    } else {
        if ( n <= 9999999 ) {
            if ( n <= 999999 ) return 6;
            else               return 7;
        } else {
            if ( n <= 99999999 ) return 8;
            else                 return 9;
        }
    }
}

The stats for 1e8 iterations (too quick to measure at 1e7)
numDigitsWithLog for 100000000 took 67 seconds
numDigitsWithDivBy10 for 100000000 took 31 seconds
num_len for 100000000 took 3 seconds

Nick · 08-11-2002

What about this one

Code:

int num_digits(int n)
{
    int count = 0;

    if (n < 10)
        return 1;
    else if (n < 100)
        return 2;

    do {
        count += 3;
        n /= 1000;
    } while(n >= 1000);

    if (n) {
        count++;
        if (n >= 10) {
            count++;
            if (n >= 100)
                count++;
        }
    }

    return count;
}

**Nick** · 08-11-2002

Still somewhat slower. You can cut out the top if statements
and make the do while into a while.

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