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C++ - how to find the length of an integer

I'm trying to find a way to find the length of an integer (number of digits) and then place it in an integer array. The assignment also calls for doing this without the use of classes from the STL, although the program spec does say we can use "common C libraries" (gonna ask my professor if I can use cmath, because I'm assuming log10(num) + 1 is the easiest way, but I was wondering if there was another way).

Ah, and this doesn't have to handle negative numbers. Solely non-negative numbers.

I'm attempting to create a variant "MyInt" class that can handle a wider range of values using a dynamic array. Any tips would be appreciated! Thanks!

13 Answers
13

The number of digits of an integer n in any base is trivially obtained by dividing until you're done:

n

unsigned int number_of_digits = 0;

do
++number_of_digits;
n /= base;
while (n);


Not necessarily the most efficient, but one of the shortest and most readable using C++:

std::to_string(num).length()


If you can use C libraries then one method would be to use sprintf, e.g.

#include <cstdio>

char s[32];

int len = sprintf(s, "%d", i);


s

"I mean the number of digits in an integer, i.e. "123" has a length of 3"

int i = 123;

// the "length" of 0 is 1:
int len = 1;

// and for numbers greater than 0:
if (i > 0)
// we count how many times it can be divided by 10:
// (how many times we can cut off the last digit until we end up with 0)
for (len = 0; i > 0; len++)
i = i / 10;



// and that's our "length":
std::cout << len;


outputs 3

3

Closed formula for the size of the int:

int

ceil(8*sizeof(int) * log10(2))


EDIT:

For the number of decimal digits of some value:

ceil(log10(var+1))


This works for numbers > 0. Zero must be checked separately.

> 0

int

ceil(log10(var+1))

int intLength(int i)
int l=0;
for(;i;i/=10) l++;
return l==0 ? 1 : l;



Here's a tiny efficient one

Being a computer nerd and not a maths nerd I'd do:

char buffer[64];
int len = sprintf(buffer, "%d", theNum);


There is a much better way to do it

#include<cmath>
...
int size = log10(num) + 1
....


works for int and decimal

How about (works also for 0 and negatives):

int digits( int x )
return ( (bool) x * (int) log10( abs( x ) ) + 1 );



Would this be an efficient approach? Converting to a string and finding the length property?

int num = 123
string strNum = to_string(num); // 123 becomes "123"
int length = strNum.length(); // length = 3
char array[3]; // or whatever you want to do with the length


Code for finding Length of int and decimal number:

#include<iostream>
#include<cmath>
using namespace std;
int main()

int len,num;
cin >> num;
len = log10(num) + 1;
cout << len << endl;
return 0;

//sample input output
/*45566
5

Process returned 0 (0x0) execution time : 3.292 s
Press any key to continue.
*/


Best way is to find using log, it works always

int len = ceil(log10(num))+1;


Plain and pretty simple function

int intlen(int i)

int j=1;
while(i>10)i=i/10;j++;
return j;



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