Reverse two numbers and obtain the reverse of the sum

Question

I am coming to SO as a last resort. Been trying to debug this code for the past 2 hours. If the question is suited to some other SE site, please do tell me before downvoting.

Here it goes:

#include <stdio.h>
#include<math.h>

int reverse(int n) {
    int count = 0, r, i;
    int k = (int)log(n * 1.0);
    for(i = k; i >= 0; i--)
    {
        r = (n % 10);
        n = (n / 10);
        count = count + (r * pow(10, k));
    }
    return count;
}

int main(void) {
     int t;
     scanf("%d", &t);

     while(t--)
     {
         int m, n, res;
         scanf("%d %d", &m, &n);
         res = reverse(m) + reverse(n);
         printf("%d", reverse(res));
     }
     return 0;
}

My objective is to get 2 numbers as input, reverse them, add the reversed numbers and then reverse the resultant as well.I have to do this for 't' test cases.

The problem: http://www.spoj.com/problems/ADDREV/

Any questions, if the code is unclear, please ask me in the comments. Thank you.

EDIT: The program gets compiled successfully. I am getting a vague output everytime. suppose the 2 numbers as input are 24 and 1, I get an output of 699998. If I try 21 and 1, I get 399998.

Answer 1

Okay, if you had properly debugged your code you would have notices strange values of k. This is because you use log which

Computes the natural (base e) logarithm of arg.

(took from linked reference, emphasis mine).

So as you are trying to obtain the 'length' of the number you should use log10 or a convertion (look at wiki about change of base for logarithms) like this: log(x)/log(10) which equal to log10(x)

And now let's look here: pow(10, k) <-- you always compute 10^k but you need 10^i, so it should be pow(10, i) instead.

Edit 1: Thanks to @DavidBowling for pointing out a bug with negative numbers.

I don't know how exactly you have to deal with negative numbers but here's one of possible solutions:

1. before computing k:

   bool isNegative = n < 0; n = abs(n);

2. Now your n is positive due to abs() returning absolute value. Go on with the same way.

3. After for loop let's see if n was negative and change count accordingly:

   if (isNegative) { count = -count; } return count;

Note: Using this solution we reverse the number itself and leave the sign as it is.

Answer 2

It looks like Yuri already found your problem, but might I suggest a shorter version of your program? It avoids using stuff like log which might be desirable.

#include <stdio.h>

int rev (int n) {
    int r = 0;
    do {
        r *= 10;
        r += n % 10;
    } while (n /= 10);
    return r;
}

int main (void) {
    int i,a,b;
    scanf("%d",&i);

    while (i--) {
        scanf("%d %d",&a,&b);
        printf("%d\n",rev(rev(a) + rev(b)));
    }       
    return 0;
}

Hopefully you can find something useful to borrow! It seems to work okay for negative numbers too.

Answer 3

Under the hood you get char string, reverse it to numeric, than reverse it to char. Since is more comfortable work with chars than let's char:

char * reverse (char *s,size_t len) //carefull it does it in place
{
   if (!len) return s; 
   char swp, *end=s+len-1;
   while(s<end)
   {
       swp =*s;
       *s++=*end;
       *end--=swp; 
   } 
   return s;
}
void get_num(char *curs)
{
     char c;
     while((c=getchar())!='\n')
      *curs++=c;
     *curs=0;
}
int main()
{
     double a,b,res;
     char sa[20],sb[20],sres[20],*curs;
     get_num( sa);
     get_num(sb);
     reverse(sa,strlen(sa));
     reverse(sb,strlen(sb));
     sscanf(sa,"%f",&a);
     sscanf(sb,"%f",&b);
     res=a+b;
     sprintf(sres,"%f",res);
     reverse(sres);
     printf(sres);

 }