How to resolve the algorithm Exponentiation operator step by step in the Elixir programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Exponentiation operator step by step in the Elixir 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 Elixir programming language
Source code in the elixir programming language
defmodule My do
def exp(x,y) when is_integer(x) and is_integer(y) and y>=0 do
IO.write("int> ") # debug test
exp_int(x,y)
end
def exp(x,y) when is_integer(y) do
IO.write("float> ") # debug test
exp_float(x,y)
end
def exp(x,y), do: (IO.write(" "); :math.pow(x,y))
defp exp_int(_,0), do: 1
defp exp_int(x,y), do: Enum.reduce(1..y, 1, fn _,acc -> x * acc end)
defp exp_float(_,y) when y==0, do: 1.0
defp exp_float(x,y) when y<0, do: 1/exp_float(x,-y)
defp exp_float(x,y), do: Enum.reduce(1..y, 1, fn _,acc -> x * acc end)
end
list = [{2,0}, {2,3}, {2,-2},
{2.0,0}, {2.0,3}, {2.0,-2},
{0.5,0}, {0.5,3}, {0.5,-2},
{-2,2}, {-2,3}, {-2.0,2}, {-2.0,3},
]
IO.puts " ___My.exp___ __:math.pow_"
Enum.each(list, fn {x,y} ->
sxy = "#{x} ** #{y}"
sexp = inspect My.exp(x,y)
spow = inspect :math.pow(x,y) # For the comparison
:io.fwrite("~10s = ~12s, ~12s~n", [sxy, sexp, spow])
end)
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