How to resolve the algorithm Stack step by step in the Go programming language

Published on 12 May 2024 09:40 PM
#Go

How to resolve the algorithm Stack step by step in the Go 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 Go programming language

==> Notes for the first code...

  • The code demonstrates the basic usage of a stack data structure implemented as a slice of integers.
  • It starts by initializing an empty slice and then appends an element to it.
  • Next, it pops an element from the stack, returning both the popped value and the modified slice.
  • Finally, it checks if the stack is empty and prints the last element of the stack.

==> Notes for the second code...

  • The code defines a custom stack data structure using a slice of interfaces ([]interface{}).
  • It provides three methods: push(), pop(), and peek() to add, remove, and access elements from the stack, respectively.
  • It also has an empty() method to check if the stack is empty.
  • In the main() function, a new stack is created and its initial state is printed.
  • Then, three elements are pushed onto the stack, and the stack is printed after each push.
  • Next, it checks if the stack is empty and prints the top value of the stack using the peek() method.
  • Finally, it pops an element from the stack and prints the popped value and the modified stack.

Source code in the go programming language

var intStack []int


intStack = append(intStack, 7)


popped, intStack = intStack[len(intStack)-1], intStack[:len(intStack)-1]


len(intStack) == 0


intStack[len(intStack)-1]


package main

import "fmt"

type stack []interface{}

func (k *stack) push(s interface{}) {
    *k = append(*k, s)
}

func (k *stack) pop() (s interface{}, ok bool) {
    if k.empty() {
        return
    }
    last := len(*k) - 1
    s = (*k)[last]
    *k = (*k)[:last]
    return s, true
}

func (k *stack) peek() (s interface{}, ok bool) {
    if k.empty() {
        return
    }
    last := len(*k) - 1
    s = (*k)[last]
    return s, true
}

func (k *stack) empty() bool {
    return len(*k) == 0
}

func main() {
    var s stack
    fmt.Println("new stack:", s)
    fmt.Println("empty?", s.empty())
    s.push(3)
    fmt.Println("push 3. stack:", s)
    fmt.Println("empty?", s.empty())
    s.push("four")
    fmt.Println(`push "four" stack:`, s)
    if top, ok := s.peek(); ok {
        fmt.Println("top value:", top)
    } else {
        fmt.Println("nothing on stack")
    }
    if popped, ok := s.pop(); ok {
        fmt.Println(popped, "popped.  stack:", s)
    } else {
        fmt.Println("nothing to pop")
    }
}

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