Question

can someone translate this pseudo code to actual c++ code

while (not the end of the input)

If the next input is a number read it and push it on the stack

else If the next input is an operator, read it

pop 2 operands off of the stack

apply the operator

push the result onto the stack

When you reach the end of the input:

if there is one number on the stack, print it else error

Answer 1

// Here is your pseudo code in c++ as per all instruction

#include <iostream>
#include <stack>
using namespace std;

int stack_solver(string str) // Here is function to solve input string
{
        int i=0,length,a,b;
        stack<int> s;            // Here we make instance s of stack predefined class in C++ library
        length=str.length();      // Here we input the length of strung via length()
        while(i<length)          // This will run upto end of the string
        {
        if(str[i]>='0'&&str[i]<='9')  // Here we check for operand
                        s.push(str[i]-'0');   // if oprand push in stack

           else if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))  // Here we check for any input other than operator and operand
            return -1;

                else                     // here else for operator
                {
            a= s.top();         //here we store the top most element in a
                        s.pop();           // Here we pop the topmost element
            b= s.top();        // here we store the next top most element in b
                        s.pop();          //Here we pop the next topmost element
            if(str[i]=='-')    // Here we check for operation to do on operand and push the result on stack
                                s.push(b-a);
                        else if(str[i]=='+')  // Here we check for operation to do on operand and push the result on stack
                                s.push(b+a);
                        else if (str[i]=='/')  // Here we check for operation to do on operand and push the result on stack
                                s.push(b/a);
                        else if(str[i]=='*')   // Here we check for operation to do on operand and push the result on stack
                                s.push(b*a);
         }
                i++;
        }
        if(s.size()==1)   // here we check for only one element in stack if true then return
        return s.top();
        else
        return -1;   // else return -1 to print the ERROR
}

int main()
{
        string str;   // Here we declare the string str
        cout<<"Enter the string of operator and operands ";   // Here we alert the user for input the string
        getline(cin,str);     // Here we input the string
        if(stack_solver(str)==-1)   // Here we check for any invalid input
        cout<<"ERROR";
    else
        cout<<stack_solver(str);   // Here we print our result if input is correct
    return 0;
}

Output for correct input:

Output for stack has left with more than one number

Output for incorrect input:

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