Tilted Forum Project Discussion Community - [c++]Large int into single #'s

Tilted Forum Project Discussion Community (https://thetfp.com/tfp/)

- Tilted Technology (https://thetfp.com/tfp/tilted-technology/)

- - [c++]Large int into single #'s (https://thetfp.com/tfp/tilted-technology/69088-c-large-int-into-single-s.html)

j0hnb

09-14-2004 05:42 AM

[c++]Large int into single #'s

How would I convert a larger int. ex: 8790278573 into single numbers so that I can find the sum of the int. I need 8790278573 to turn into 8 7 9 0 2 7 8 5 7 3, so that I can add them together and find the sum. Any ideas on how I should do that?

welshbyte

09-14-2004 07:52 AM

I've knocked together a little solution which works with int's but i'm afraid 8790278573 doesnt fit in an int (on my computer at least). You can see the method i used anyway:

Code:

#include<iostream>



using namespace std;



int main() {



        int i = 790278573;  // Number you want to add the digits of



        int icopy = i;

        int result = 0;

        

        while(icopy > 0) {

                result += icopy % 10;

                icopy /= 10;

        }



        cout << "Initial number: " << i      << "\n";

        cout << "Sum of digits : " << result << "\n";

}

I'd like to see a better method if anyone would like to post one. I'm not that hot on C++

Hope this helps anyway

bacon

09-14-2004 09:06 AM

Store the large number as a string. By "string" I don't mean use a c++ "String" class, use a character array. Once you have the bigbad number as a char*, loop over it character by character, and convert each back to a int.

welshbyte

09-14-2004 10:03 AM

Quote:

Originally Posted by welshbyte

i'm afraid 8790278573 doesnt fit in an int (on my computer at least).

Hmm, it should do but somethings wrong on my PC... i think i'm gonna need to look into it.

Anyway, my solution should work fine for that big number. I still cant think of a better way to do it...

bacon

09-14-2004 05:33 PM

I swear, you guys are making this way too hard.

Code:

int main()

{



        char *strBignum = "790278573\0";

        int i,sum=0;

        for(i=0; i<strlen(strBignum); i++) {

                sum += (int)(strBignum[i]-'0');

        }

        cout << "The sum is: " << sum <<endl;



}

The only trick is the ascii math. Just subtract the ascii value of '0' from the char, and cast it to an int.

roboshark

09-14-2004 10:08 PM

That's a good one, bacon. The hardest part is getting the ascii math right.

You can even make it simpler (I like to make things as simple as possible):

Code:

int main()

{

        char *p = "8790278573";

        int s = 0;



        while (*p)

                s += *p++ - '0';



        cout << "sum " << s << endl;

}

welshbyte

09-14-2004 11:12 PM

Nice one roboshark, very elegant. Have to say i was trying to stick to ints because j0hnb specified ints in his question but yours seems a much better way of doing it.

roboshark

09-15-2004 02:04 AM

Quote:

Originally Posted by welshbyte

Nice one roboshark, very elegant. Have to say i was trying to stick to ints because j0hnb specified ints in his question but yours seems a much better way of doing it.

Thanks. Elegant code is so much more rewarding.

However, don't knock your solution completely. If the input can be represented as a string (such as reading the number from the command line), a solution such as bacon's or mine would be the best one. However, if the input is already in integer form (though the integer the OP specified won't fit into a 32 bit integer), then a solution along the lines you've presented will be the only choice. And yours looks pretty elegant to me!

bacon

09-15-2004 05:07 AM

Pointer arithmatic is swanky. We can tweak this a bit more.

Code:



int main()

{

        int s = 0;

        for (char *p = "8790278573"; *p; s+= *p++ -'0');

        cout << "sum " << s << endl;

}

I hate while loops. :)

welshbyte

09-15-2004 09:04 AM

robo: thanks for the kind words :)

You know, If we tweak this any more its gonna be totally unreadable!

Heh, i never knew you could go to the next character in a *char by using ++. Ya learn something new every day :)

Anyway, i think j0hnb should have his answer now *pats us all on the back* I wonder where he went...

Rekna

09-15-2004 04:40 PM

If your original data is in an integer format go with the first solution. Also the first solution can be modified slightly to change your base. You should learn the mod divide algorithm for extracting digits.

oblar

09-15-2004 11:03 PM

you guys are my heros :) hanging around a bunch of java nuts makes me miss pointers soo much :)

All times are GMT -8. The time now is 01:14 AM.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62