Thread: add one integer with is own number

Anyone have an easy way to do that

Code:

int number  = 18;
int number2 = 2;
int addnumber;

//I want to add 1 + 8 + 2 for example so that int addnumber = 11

by the way, it's not sure that the number will be over 10 so there is surely a if
thanks once again

I'm not sure your math background but you could check the number of digits with log(|number|). As far as solving your overall problem, I think the easiest(most straightforward) way would be to convert the numbers to a string and then add each character of the string to the total. You can do that or do some pure math with something like:

Code:

// this will add all the digits in number
int digit_sum = 0;
for (int i = 0; i <= (int)log(abs(number)); i++)
	digit_sum += (int)(abs(number) / pow(10, i)) &#37; 10;

Code:

(int)log(abs(number))

this returns the number of digits in the number

Code:

(int)(abs(number) / pow(10, i)) % 10)) % 10

This singles out the ith digit. For example, in the number 486, 6 is the 0th digit, 8 is the 1st digit, 4 is the 2nd digit.

mmmm....something like this should work:

Code:

num=0x000345
int digit,digit_total;
//get rid of the leading zeros
int i =0;
while(!(digit=(num<<i)&(1<<sizeof(int))) && i < sizeof(int))
   i++;

//now count
digit_total = 0
while ( i < sizeof(int) )
{
     digit = ( num << i ) & ( 1 <<sizeof(int));
     digit_total +=digit;
}

There's no need to check the number of digits.

Code:

int sumdig( int d )
{
    int x = 0;
    while( d )
    {
        x += d % 10;
        d /= 10;
    }
    return x;
}

Quzah.

oops !. Disregard my previous reply. I was thinking about binary number