count the digits of a number with recursion

Question

#include<iostream>
using namespace std;

bool recursion(int numb, int k, int br)
{
    if(br==1) return (numb==k);
    return k==(numb%10) || recursion(numb/10,k,br-1);

}

int main(){
    int num,n;
    cin>>num;
    n=num;
    int p;
    cin>>p;
    int br=1;
    while(n>10){
        n=n/10;
        br++;
    }
    cout<<br<<endl;
    cout<<recursion(num,p,br);
    return 0;
}

This is the whole program for counting the digits of a number , but it doesn't work for numbers with more than 10 digits. Does anybody know why?

Try `long long` instead of `int` if you have a supporting ocmpiler. — Shiv, Jan 24 '15 at 12:30
there is a limit to (int) please see this answer http://stackoverflow.com/questions/1819189/what-range-of-values-can-integer-types-store-in-c — Mouhamad Arafeh, Jan 24 '15 at 12:34

score 1 · Answer 1 · answered Jan 24 '15 at 12:29

First, your recursive program is not counting the digits in a number, it checks if a particular digit k is present within the last br digits of the number numb.

It does not work for numbers with more than ten digits because the largest number on your system that int can represent has ten digits. On 32-bit systems it is 2,147,483,647.

To make it work with more digits, use a larger data type - say, long long, or uint64_t.

score 0 · Answer 2 · answered Jan 24 '15 at 12:30

0

On a 32-bits machine, integers are 32 bits long. The largest signed integer you can get is 2^31 - 1 = 2147483647 which has 10 digits. You've got to use strings to allow for arbitrarily large numbers.

answered Jan 24 '15 at 12:30

mrk

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no they aren't necessary. The length of `int` is implementation defined. For example, on my 64 bit machine, `int`s are also 32 bit. – vsoftco Jan 24 '15 at 13:18

count the digits of a number with recursion

2 Answers2