Step by Step solution about How to resolve the algorithm Exponentiation operator step by step in the C programming language

The C code provided defines two different functions to calculate the power of a number. The first function, ipow(), is used to calculate the integer power of an integer base. The second function, dpow(), is used to calculate the double-precision floating-point power of a double-precision floating-point base. Additionally, the code includes a generic generic_pow() function that can be used to calculate the power of either an integer or a double-precision floating-point number based on the type of the base argument.

Here's a detailed explanation of the code:

• ipow() Function:

  • This function takes two integer arguments: base and exp. It calculates the integer power of base raised to the power of exp.
  • It uses an iterative approach to calculate the power. It initializes a variable pow to the value of base and a variable v to 1.
  • It checks if exp is negative. If exp is negative, it asserts that base is not 0 (to avoid division by zero) and returns 0 if base*base != 1 (i.e., base is odd), and returns base if exp is odd, or 1 if exp is even.
  • It iterates until exp becomes 0. In each iteration, it multiplies v by pow if the least significant bit of exp is 1. It then multiplies pow by itself and shifts exp right by 1 bit.
  • Finally, it returns the value of v, which is the result of base raised to the power of exp.

• dpow() Function:

  • This function takes two double-precision floating-point arguments: base and exp. It calculates the double-precision floating-point power of base raised to the power of exp.
  • It uses a similar iterative approach as the ipow() function. It initializes a variable pow to the value of base (or 1.0 if exp is negative) and a variable v to 1.0.
  • It checks if exp is negative. If exp is negative, it makes exp positive.
  • It iterates until exp becomes 0. In each iteration, it multiplies v by pow if the least significant bit of exp is 1. It then multiplies pow by itself and shifts exp right by 1 bit.
  • Finally, it returns the value of v, which is the result of base raised to the power of exp.

• generic_pow() Macro:

  • This macro uses the C language's _Generic() feature to select the appropriate function to call based on the type of the base argument.
  • It takes two arguments: base and exp.
  • If the type of base is double, it calls the dpow() function with the given arguments.
  • If the type of base is int, it calls the ipow() function with the given arguments.
  • It returns the result of the function call.

• Main Function:

  • The main function demonstrates the usage of the ipow(), dpow(), and generic_pow() functions.
  • It prints the result of 2^6, 2^-6, 2.71^6, and 2.71^-6 using both the ipow()/dpow() functions and the generic_pow() macro.

#include <stdio.h>
#include <assert.h>

int ipow(int base, int exp)
{
   int pow = base;
   int v = 1;
   if (exp < 0) {
      assert (base != 0);  /* divide by zero */
      return (base*base != 1)? 0: (exp&1)? base : 1;
   }
       
   while(exp > 0 )
   {
      if (exp & 1) v *= pow;
      pow *= pow;
      exp >>= 1;
    }
   return v;
}

double dpow(double base, int exp)
{
   double v=1.0;
   double pow = (exp <0)? 1.0/base : base;
   if (exp < 0) exp = - exp;

   while(exp > 0 )
   {
      if (exp & 1) v *= pow;
      pow *= pow;
      exp >>= 1;
   }
   return v;
}

int main()
{
    printf("2^6 = %d\n", ipow(2,6));
    printf("2^-6 = %d\n", ipow(2,-6));
    printf("2.71^6 = %lf\n", dpow(2.71,6));
    printf("2.71^-6 = %lf\n", dpow(2.71,-6));
}


#define generic_pow(base, exp)\
    _Generic((base),\
            double: dpow,\
            int: ipow)\
    (base, exp)

int main()
{
    printf("2^6 = %d\n", generic_pow(2,6));
    printf("2^-6 = %d\n", generic_pow(2,-6));
    printf("2.71^6 = %lf\n", generic_pow(2.71,6));
    printf("2.71^-6 = %lf\n", generic_pow(2.71,-6));
}