How to resolve the algorithm Stack step by step in the M2000 Interpreter programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Stack step by step in the M2000 Interpreter programming language

Table of Contents

Problem Statement

A stack is a container of elements with   last in, first out   access policy.   Sometimes it also called LIFO. The stack is accessed through its top. The basic stack operations are:

Sometimes the last pushed stack element is made accessible for immutable access (for read) or mutable access (for write):

Stacks allow a very simple hardware implementation. They are common in almost all processors. In programming, stacks are also very popular for their way (LIFO) of resource management, usually memory. Nested scopes of language objects are naturally implemented by a stack (sometimes by multiple stacks). This is a classical way to implement local variables of a re-entrant or recursive subprogram. Stacks are also used to describe a formal computational framework. See stack machine. Many algorithms in pattern matching, compiler construction (e.g. recursive descent parsers), and machine learning (e.g. based on tree traversal) have a natural representation in terms of stacks.

Create a stack supporting the basic operations: push, pop, empty.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Stack step by step in the M2000 Interpreter programming language

Source code in the m2000 programming language

Module Checkit {
      a=Stack
      Stack a {
            Push 100, 200, 300
      }
      Print StackItem(a, 1)=300
      Stack a {
            Print StackItem(1)=300
            While not empty {
                  Read N
                  Print N
            }
      }
}
Checkit

Module Checkit {
      Read a, b
      Print a, b
}
\\ add parameters in a FIFO, and this FIFO merged to current stack
Push 100
Checkit 10, 20
Print StackItem(1)=100
Module Checkit {
      Read a, b
      Print a=20, b=100
}
Checkit 20

Function alfa {
      k=0
      n=0
      while not empty {
            k+=number
            n++
      }
      if n=0 then Error "No parameters found"
      =k/n
}

Print alfa(1,2,3,4)=2.5

  

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