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

问题

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!

回答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);

回答2:

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

std::to_string(num).length()

回答3:

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);

回答4:

"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

回答5:

There is a much better way to do it

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

works for int and decimal

回答6:

Closed formula for the size of the 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.

回答7:

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

Here's a tiny efficient one

回答8:

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

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

回答9:

How about (works also for 0 and negatives):

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

回答10:

Best way is to find using log, it works always

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

回答11:

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

回答12:

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.
    */

回答13:

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;
}

来源：https://stackoverflow.com/questions/22648978/c-how-to-find-the-length-of-an-integer

标签

c++

count

numbers

integer

digits