How to resolve the algorithm Exponentiation operator step by step in the FreeBASIC programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Exponentiation operator step by step in the FreeBASIC 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 FreeBASIC programming language
Source code in the freebasic programming language
' FB 1.05.0
' Note that 'base' is a keyword in FB, so we use 'base_' instead as a parameter
Function Pow Overload (base_ As Double, exponent As Integer) As Double
If exponent = 0.0 Then Return 1.0
If exponent = 1.0 Then Return base_
If exponent < 0.0 Then Return 1.0 / Pow(base_, -exponent)
Dim power As Double = base_
For i As Integer = 2 To exponent
power *= base_
Next
Return power
End Function
Function Pow Overload(base_ As Integer, exponent As Integer) As Double
Return Pow(CDbl(base_), exponent)
End Function
' check results of these functions using FB's built in '^' operator
Print "Pow(2, 2) = "; Pow(2, 2)
Print "Pow(2.5, 2) = "; Pow(2.5, 2)
Print "Pow(2, -3) = "; Pow(2, -3)
Print "Pow(1.78, 3) = "; Pow(1.78, 3)
Print
Print "2 ^ 2 = "; 2 ^ 2
Print "2.5 ^ 2 = "; 2.5 ^ 2
Print "2 ^ -3 = "; 2 ^ -3
Print "1.78 ^ 3 = "; 1.78 ^ 3
Print
Print "Press any key to quit"
Sleep
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