Question

Implement the array-based stack class. Use it in the client code to convert and compute post-fix expressions. The postfix expressions are saved in a file named "postfix.txt", one expression each line. For simplicity, you may assume all numbers in the expressions are single digit numbers.

Submit:
1. Source code: the stack class and the client code;
2. Output screen capture: show the results for the following expressions (The errors in some expressions are intentional. Your code must detect it and output an error message):

3 7 +

3 7 2 * +

3 7 + 2 *

* 3 7 2 4 - * 5 + 2 * 9 + 4 8 - * + 1 –

3 7 2 4 - * 5 + 2 * 9 + 4 8 - * + 1 - +

3 7 2 4 - * 5 + * 2 * 9 + 4 8 - * + 1 –

3 7 2 4 - * 5 + 2 * 9 + 4 8 - * + 1 -

All files must be submitted individually. DO NOT ZIP.

Answer 1

//C++ program

#include <iostream>

using namespace std;

class stack{

int arr;

int top;

int capacity;

public:

stack(int size){

capacity = size;

arr = new int[capacity];

top=-1;

}

bool isempty(){

return top==-1;

}

void push (int val){

arr[++top]=val;

}

void pop (){

top--;

}

int top(){

return arr[top];

}

};

int main(){

int n,num1,num2;

cout<<"Enter length of expression : ";

cin>>n;

char a[n];

cout<<"Enter expression : ";

for (int i=0;i<n;i++){

cin>>a[i];

}

stack s(n);

for(int i=0;i<n ;i++){

if(a[i]>='0' &&a[i]<='9') s.push(a[i]-'0');

else{

num2=s.top();

s.pop();

num1 = s._top();

s.pop();

if(a[i] == '+')s.push(num1 + num2);

if(a[i] == '-')s.push(num1 - num2);

if(a[i] == '*')s.push(num1 num2);

if(a[i] == '/')s.push(num1 / num2);

}

}

cout<<"\nvalue of expression : "<< s._top()<<"\n";

return 0;

}

//sample output $CAUsers\IshuManish\Documents\ poste.exe nter length o expression5 nter expression 3 7 2 * + value of expression 17 alue of expressionl? Process exited after 21.86 seconds with return value 0 Press any key to continue - .$