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!
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!
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
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
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 ‘0’ 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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