STACK PROJECT e a program that uses stacks to evaluate an arithmetic expression in infix notation without converting postfix notation. program takes as input a numeric expression in infix notation, such as 3+4*2, and outputs the result. Operators are +, -, *, / Assume that the expression is formed correctly so that each operation has two arguments. he expression can have parenthesis, for example: 3*(4-2)+6. he expression can have negative numbers. he expression can have spaces in it, for example: 3 * (4-2) +6. are some useful functions that you may need: char cin.peek(); -- returns the next character of the cin input stream ( without reading it) beel isdigitle digits O' thrGush (O!

C++ for Engineers and Scientists
4th Edition
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter5: Repetition Statements
Section: Chapter Questions
Problem 7PP
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Question

Given:

1.Stack.d file

#include <iostream>

using namespace std;

#ifndef STACK_H
#define STACK_H

template <class T>
struct node
{
T info;
node *next;
};

template <class T>
class Stack
{
private:
node<T> *top;
 
public:
Stack();
~Stack();
Stack ( const Stack<T> & other);
const Stack<T> & operator= (const Stack<T> & other);
bool isEmpty();
void push(T item);
void pop();
int getLength();
void Top(T&);
void copy (const Stack<T> & other);
void destroy();
};
#endif
 
2. Stack.cpp
 
#include "Stack.d"
template <class T>
Stack<T> :: Stack ()
{
top = NULL;
}

template <class T>
Stack<T> :: ~Stack()
{
destroy();
}

template <class T>
Stack<T> :: Stack ( const Stack<T> & other )
{
copy (other);
}

template <class T>
const Stack<T> & Stack<T> :: operator= ( const Stack<T> & other )
{
if ( this != &other )
{
destroy();
copy (other);
}
return *this;
}

template <class T>
bool Stack<T> :: isEmpty ()
{
return top == NULL;
}

template <class T>
void Stack<T> :: push ( T item )
{
node<T> *temp;

temp = new node<T>;      
temp->info = item;     
temp->next = top;      
top = temp;      
}

template <class T>
void Stack<T> :: pop ()
{
if (!isEmpty ())
{
node<T> *temp;

temp = top;
top = top->next;
delete temp;
}
}

template <class T>
int Stack<T> :: getLength()
{
int count = 0;
node<T> *p = top;

while ( p != NULL )
{
count++;
p = p->next;
}

return count;
}

template <class T>
void Stack<T> :: Top (T &item)
{
if (!isEmpty())
item = top->info;
}

template <class T>
void Stack<T> :: copy (const Stack<T> & other)
{
if ( other.first == NULL )
top = NULL;
else
{
top = new node<T>;
top->info = other.first->info;

node<T> *p = other.top->next;
node<T> *r = top;

while ( p != NULL )
{
r->next = new node<T>;
r->next->info = p->info;
p = p->next;
r = r->next;
}
 
r->next = NULL;
}
}

template <class T>
void Stack<T> :: destroy()
{
node<T> *p;

while ( top != NULL )
{
p = top;
top = top->next;
delete p;
}
}
Finish just the main function based on the picture using the given code
STACK PROJECT
Write a program that uses stacks to evaluate an arithmetic expression in infix notation without converting it
into postfix notation.
The program takes as input a numeric expression in infix notation, such as 3+4*2, and outputs the result.
1) Operators are +, -, *, /
2) Assume that the expression is formed correctly so that each operation has two arguments.
3) The expression can have parenthesis, for example: 3*(4-2)+6.
4) The expression can have negative numbers.
5) The expression can have spaces in it, for example: 3 *
(4-2) +6.
Here are some useful functions that you may need:
char cin.peek(); -- returns the next character of the cin input stream ( without reading it)
bool isdigit(char c); -- returns true if c is one of the digits 'O' through '9', false otherwise
cin.ignore(); -- reads and discards the next character from the cin input stream
cin.get(char &c); -- reads a character in c ( could be a space or the new line )
Transcribed Image Text:STACK PROJECT Write a program that uses stacks to evaluate an arithmetic expression in infix notation without converting it into postfix notation. The program takes as input a numeric expression in infix notation, such as 3+4*2, and outputs the result. 1) Operators are +, -, *, / 2) Assume that the expression is formed correctly so that each operation has two arguments. 3) The expression can have parenthesis, for example: 3*(4-2)+6. 4) The expression can have negative numbers. 5) The expression can have spaces in it, for example: 3 * (4-2) +6. Here are some useful functions that you may need: char cin.peek(); -- returns the next character of the cin input stream ( without reading it) bool isdigit(char c); -- returns true if c is one of the digits 'O' through '9', false otherwise cin.ignore(); -- reads and discards the next character from the cin input stream cin.get(char &c); -- reads a character in c ( could be a space or the new line )
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