Can't raise x to the power of y

[xp5]

Hi, is there a way to raise x to the power of y (x^y), without using the math.h library and the pow(x,y) function?

[Happy_Reaper]

With a for loop ?

[MacGyver]

Write pow() yourself.

I don't think you want to do that. Just use pow() if you can.

[xp5]

yeah, i have to do it without using pow(x,y)

[MacGyver]

So write your own pow() function.

[Rashakil Fol]

Use a for loop. Or use a for loop that squares and loops through the bits of y.

[xp5]

Hi I'm trying to use the for loop, but i'm having trouble figuring out how to use it. Do you know what i'm doing wrong?

  int n;
  int output[60];
  int index = 0;
  int condition;
  int quotient;
  int size = 0;
  int nsize;
  int converted_number=0;
  int power;
  int i;

  printf("Enter base(b) between [1,10], and a number(n), such that digits of n is between[0,b-1] in this format 'n b': ");
  scanf("%d %d", &n, &b);

  printf("n is: %d\n", n);   //Test
  printf("b is: %d\n", b);   //TEST
  nsize = n;

  if ((b < 1) || (b > 10))
  {
    printf("Your base is not between 1 and 10");
    return 0;
  }

  if (n < 0)
  {
    printf("n must be positive");
    return 0;
  }

  while (nsize != 0)  //counting the total number of digits
  {
    nsize = nsize/10;
    ++size;
  }
  printf("size of n is: %d \n", size);  //testing to see the size

  while (size > 0)
  {
    output[index] = n % 10;
    n = n/10;
    printf("number to convert: %d\n", output[index]);

    for (i = 1; i <=index; i++)       //FOR LOOP IN HERE!!!
    {
      power = output[index]*;      //NOT SURE ON WHAT TO DO HERE
      converted_number += (power*output[index]);
    }

    printf("%d * %d^%d %d\n", output[index], b, index, converted_number);
    --size;
    index++;
  }

  printf("converted number is: %d", converted_number);



  return 0;
}

[ssharish2005]

int power(int base, int ext)
{ 
   int i;
   int res=1;
   
   for(i=1; i <= ext; i++)
      res = res * base;

   return res;
}

here is a basic pow function. It can handle just int. You will have to work around to make it to work for all datatype.

ssharish2005

[zacs7]

x^y simply means, mutliply x by itself y times.

Such that, 5^10
= 5 * 5 * 5 * 5 *5 * 5 * 5 * 5 * 5 *5

Quite easy to put in a loop.

[robatino]

This sounds like a homework problem, but I'll give you the hint that pow(x, y) can be rewritten in terms of other functions from <math.h>, so that's one way, if you're allowed to use those. Otherwise, if y is an integer, you can implement it in terms of a loop as described above.

[iMalc]

Oh, you mean it only has to work for integer values of y?
Well I wont bother to post my solution then.