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

Question

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

Answer 1

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

unsigned int number_of_digits = 0;

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

Answer 2

There is a much better way to do it

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

works for int and decimal

Answer 3

Closed formula for the longest int (I used int here, but works for any signed integral type):

1 + (int) ceil((8*sizeof(int)-1) * log10(2))

Explanation:

                  sizeof(int)                 // number bytes in int
                8*sizeof(int)                 // number of binary digits (bits)
                8*sizeof(int)-1               // discount one bit for the negatives
               (8*sizeof(int)-1) * log10(2)   // convert to decimal, because:
                                              // 1 bit == log10(2) decimal digits
    (int) ceil((8*sizeof(int)-1) * log10(2))  // round up to whole digits
1 + (int) ceil((8*sizeof(int)-1) * log10(2))  // make room for the minus sign

For an int type of 4 bytes, the result is 11. An example of 4 bytes int with 11 decimal digits is: "-2147483648".

If you want the number of decimal digits of some int value, you can use the following function:

unsigned base10_size(int value)
{
    if(value == 0) {
        return 1u;
    }

    unsigned ret;
    double dval;
    if(value > 0) {
        ret = 0;
        dval = value;
    } else {
        // Make room for the minus sign, and proceed as if positive.
        ret = 1;
        dval = -double(value);
    }

    ret += ceil(log10(dval+1.0));

    return ret;
}

I tested this function for the whole range of int in g++ 9.3.0 for x86-64.

Answer 4

Most efficient code to find length of a number.. counts zeros as well, note "n" is the number to be given.

#include <iostream>
using namespace std;
int main()
{
    int n,len= 0;
    cin>>n;
while(n!=0)
    {
       len++;
       n=n/10;
    }
    cout<<len<<endl;
    return 0;
}

Answer 5

"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

Answer 6

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.
    */