How to resolve the algorithm Greatest element of a list step by step in the Oberon-2 programming language

Published on 12 May 2024 09:40 PM
#Oberon-2

How to resolve the algorithm Greatest element of a list step by step in the Oberon-2 programming language

Table of Contents

Problem Statement

Create a function that returns the maximum value in a provided set of values, where the number of values may not be known until run-time.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Greatest element of a list step by step in the Oberon-2 programming language

Source code in the oberon-2 programming language

MODULE GreatestElement1;
IMPORT
  ADT:ArrayList,
  Object:Boxed,
  Out;
  
VAR
  a: ArrayList.ArrayList(Boxed.LongInt);
  max: Boxed.LongInt;
  
  
  PROCEDURE Max(al: ArrayList.ArrayList(Boxed.LongInt)): Boxed.LongInt;
  VAR
    i: LONGINT;
    item, max: Boxed.LongInt;
  BEGIN
    max := NEW(Boxed.LongInt,MIN(LONGINT));
    
    i := 0;
    WHILE (i < al.size) DO
      item := al.Get(i);
      IF item.value > max.value THEN max := item END;
      INC(i)  
    END;
    RETURN max
  END Max;
  
BEGIN
  a := NEW(ArrayList.ArrayList(Boxed.LongInt),5);
  a.Append(NEW(Boxed.LongInt,10));
  a.Append(NEW(Boxed.LongInt,32));
  a.Append(NEW(Boxed.LongInt,4));
  a.Append(NEW(Boxed.LongInt,43));
  a.Append(NEW(Boxed.LongInt,9));
  
  max := Max(a);
  Out.String("Max: ");Out.LongInt(max.value,4);Out.Ln
END GreatestElement1.

MODULE GreatestElement2;
IMPORT
  Out;
VAR
  a: ARRAY 10 OF LONGINT;
  
  PROCEDURE Max(a: ARRAY OF LONGINT): LONGINT;
  VAR
    i, max: LONGINT;
  BEGIN
    max := MIN(LONGINT);
    FOR i := 0 TO LEN(a) - 1 DO
      IF a[i] > max THEN max := a[i] END;
    END;
    
    RETURN max
  END Max;
BEGIN
  a[0] := 10;
  a[1] := 32;
  a[2] := 4;
  a[3] := 43;
  a[4] := 9;
  
  Out.String("Max: ");Out.LongInt(Max(a),4);Out.Ln
END GreatestElement2.

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