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Algorithm for extracting an integers digits?

Algorithm for extracting an integers digits?

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seuzy13	18 Jul, 2008 - 07:38 PM Post #1
New D.I.C Head Joined: 18 Jul, 2008 Posts: 4	Hello! I've been trying to write a library function that will extract a specified digit from an integer. I've always hated trying to find the digits of a number, because just thinking about all the dividing and modulusing makes me want to weep softly in a corner. I did manage to write a little program once that would separate a five digit integer's digits by spaces, but I'm thinking it only works for five digits numbers? Anyway, I only got it to work by fiddling with a million different possibilities, so a clear explanation would be appreciated. I did a google search on this and came up with nothing. Thanks for your help! The declaration of the digit extracting function: CODE int digit (int num, int digit); //finds the digit of a number; digits are counted from right to left, so for example, the hundreds place //would be specified using the digit number 3 The program I mentioned: CODE //1-36.cpp //Puts spaces between the digits of a user-inputted integer; //Author-SB #include <iostream> using namespace std; int main () { int num; //user-inputted integer to get digits of /~get numbers~/ cout << "Enter a five-digit number: "; cin >> num; /~determine and output digits~/ cout << num / 10000 << " " << num / 1000 % 10 << " " << num / 100 % 100 % 10 << " " << num / 10 % 1000 % 100 % 10 << " " << num % 10000 % 1000 % 100 % 10<< endl; /note to self: there is a definite algorithm to finding digits, and I should definitely write a library function to handle this commonly occuring problem, because the algorithm that does exist makes me want to weep softly in a corner/ return 0; //return successfully }
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NickDMax	18 Jul, 2008 - 09:01 PM Post #2
2B\|\|!2B Joined: 18 Feb, 2007 Posts: 2,672 Thanked 26 times Dream Kudos: 475 My Contributions	to extract a given digit: take the number and divide it by 10^(digit -1), then find result mod 10. You need to do a couple of checks: First off, if digit == 1 then 10^(digit-1) == 0 so you would divide by zero (bad -- end of the universe and all). So for this you need a seperate calculation of finding the mod 10 of the number. Secondly, barring the above digit should be between 2 and log10(number) + 1 (the total number of digits). The upper limit is not really all that important since it will just generate a zero, though if digit gets too large then you will cause an overflow.
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seuzy13	19 Jul, 2008 - 08:08 AM Post #3
New D.I.C Head Joined: 18 Jul, 2008 Posts: 4	QUOTE(NickDMax @ 18 Jul, 2008 - 09:01 PM) to extract a given digit: take the number and divide it by 10^(digit -1), then find result mod 10. You need to do a couple of checks: First off, if digit == 1 then 10^(digit-1) == 0 so you would divide by zero (bad -- end of the universe and all). So for this you need a seperate calculation of finding the mod 10 of the number. Secondly, barring the above digit should be between 2 and log10(number) + 1 (the total number of digits). The upper limit is not really all that important since it will just generate a zero, though if digit gets too large then you will cause an overflow. Thanks, that seems to work, except its a sort of "off by one error." For example, I have the arbitrary number 546721. When I ask for the 1000s digit I get 4. CODE result=(number / ( 10^(digit-1) ) )%10; Also, there is no problem with dividing by zero, because if the digit is one, you get ten to the "zeroth" power. Anything to the zeroth power is one, so there would be no problem with that. Or does it happen to work differently in C++? This post has been edited by seuzy13: 19 Jul, 2008 - 08:20 AM
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NickDMax

19 Jul, 2008 - 08:39 AM

Post #4

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no it does not work differently in C++ that was just late night stupidity on my part.

However 10^(digit-1)... well it turns out the the ^ operator is XOR in C++...

this is what I came up with:

cpp

#include <iostream>
#include <cmath>
using namespace std;

int getDigit(unsigned int number, unsigned int digit);

int main() {
    unsigned int num = 123456789;
    cout << "The number is: " << num << endl;
    for(int i = 0; i < 12; i++) {
        cout << "Digit " << i+1 << " = " << getDigit(num,(i+1)) << "\n";
    }
    cout << endl;
    return 0;
}

int getDigit(const unsigned int number, const unsigned int digit) {
    if (digit > 0 && digit <= log10(number)+1) {
       return (int)(number / pow((double)10, (double)(digit-1))) % 10;
    }  else {
        return 0;
    }
}

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seuzy13	19 Jul, 2008 - 08:49 AM Post #5
New D.I.C Head Joined: 18 Jul, 2008 Posts: 4	Haha, cheers to late night stupidity! Thanks, this works great. Now I can add it to my library and never think about the modulus operator again.
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NickDMax	19 Jul, 2008 - 09:53 AM Post #6
2B\|\|!2B Joined: 18 Feb, 2007 Posts: 2,672 Thanked 26 times Dream Kudos: 475 My Contributions	Don't you wish!!! The modulus operator is actually amazingly handy.
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seuzy13	19 Jul, 2008 - 10:33 AM Post #7
New D.I.C Head Joined: 18 Jul, 2008 Posts: 4	QUOTE(NickDMax @ 19 Jul, 2008 - 09:53 AM) Don't you wish!!! The modulus operator is actually amazingly handy. Yeah, I know. But at least I'm off the hook for this problem.
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baavgai	19 Jul, 2008 - 06:04 PM Post #8
Dreaming Coder Joined: 16 Oct, 2007 Posts: 1,471 Thanked 38 times Dream Kudos: 325 Expert In: C, C++, Java, C#, ASP.NET, PHP, Perl, Python, Oracle, SQL Server, MySql, HTML, JavaScript, Lua My Contributions	The answer offered is the same methodology I'd use. However, the two heavy hitting math functions of log10 and pow aren't needed. This is a far more basic approach. cpp int getDigit(unsigned int number, int digit) { for(int i=0; i<digit-1; i++) { number /= 10; } return number % 10; } Note, you really should understand how modulus works. It's a good tool and you'll be exposed to again.
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