How to resolve the algorithm Exponentiation operator step by step in the Nim programming language

Published on 12 May 2024 09:40 PM
#Nim

How to resolve the algorithm Exponentiation operator step by step in the Nim programming language

Table of Contents

Problem Statement

Most programming languages have a built-in implementation of exponentiation.

Re-implement integer exponentiation for both   intint   and   floatint   as both a procedure,   and an operator (if your language supports operator definition). If the language supports operator (or procedure) overloading, then an overloaded form should be provided for both   intint   and   floatint   variants.

Let's start with the solution:

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

Source code in the nim programming language

proc `^`[T: float|int](base: T; exp: int): T =
  var (base, exp) = (base, exp)
  result = 1

  if exp < 0:
    when T is int:
      if base * base != 1: return 0
      elif (exp and 1) == 0: return 1
      else: return base
    else:
      base = 1.0 / base
      exp = -exp

  while exp != 0:
    if (exp and 1) != 0:
      result *= base
    exp = exp shr 1
    base *= base

echo "2^6 = ", 2^6
echo "2^-6 = ", 2 ^ -6
echo "2.71^6 = ", 2.71^6
echo "2.71^-6 = ", 2.71 ^ -6

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