How to resolve the algorithm Exponentiation operator step by step in the Seed7 programming language

Published on 12 May 2024 09:40 PM
#Seed7

How to resolve the algorithm Exponentiation operator step by step in the Seed7 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 Seed7 programming language

Source code in the seed7 programming language

const func integer: intPow (in var integer: base, in var integer: exponent) is func
  result
    var integer: result is 0;
  begin
    if exponent < 0 then
      raise(NUMERIC_ERROR);
    else
      if odd(exponent) then
        result := base;
      else
        result := 1;
      end if;
      exponent := exponent div 2;
      while exponent <> 0 do
        base *:= base;
        if odd(exponent) then
          result *:= base;
        end if;
        exponent := exponent div 2;
      end while;
    end if;
  end func;

const func float: fltIPow (in var float: base, in var integer: exponent) is func
  result
    var float: power is 1.0;
  local
    var integer: stop is 0;
  begin
    if base = 0.0 then
      if exponent < 0 then
        power := Infinity;
      elsif exponent > 0 then
        power := 0.0;
      end if;
    else
      if exponent < 0 then
        stop := -1;
      end if;
      if odd(exponent) then
        power := base;
      end if;
      exponent >>:= 1;
      while exponent <> stop do
        base *:= base;
        if odd(exponent) then
          power *:= base;
        end if;
        exponent >>:= 1;
      end while;
      if stop = -1 then
        power := 1.0 / power;
      end if;
    end if;
  end func;

$ syntax expr: .(). ^* .() is <- 4;

const func integer: (in var integer: base) ^* (in var integer: exponent) is
  return intPow(base, exponent);

const func float: (in var float: base) ^* (in var integer: exponent) is
  return fltIPow(base, exponent);

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