Question

Write a C++ program that reads string expressions from an input file "prefix.txt" containing prefix expressions (one expression per input line). For each prefix expression read from the input file, the program should: (1) convert the expression to a fully parenthesized infix expression and (2) find the value of the expression. Use a stack to solve the problem. Assume the expressions contain only integer numbers and the *,/, +, - operators. Generate the results in an output file "results.txt" in the form of a table with 3 columns as shown in the example below. Example of an input file "prefix.txt": The obtained results.txt" output file for the above example of input file:

Answer 1

#include <algorithm>
#include <deque>
#include <iostream>
#include <string>

typedef std::deque< std::string > tokens_t;

void pre_to_in( tokens_t* eq ) 
{
    if ( !eq->empty() ) {
        const std::string token = eq->front();
        eq->pop_front();
        if ( ( token == "+" ) || ( token == "-" ) || ( token == "/" ) || ( token == "*" ) ) {
            std::cout << "(";
            pre_to_in( eq );
            std::cout << token;
            pre_to_in( eq );
            std::cout << ")";
        } else {
            std::cout << token;
        }
    }   
} 
using namespace std;
struct node
{
    int data;
    node *next;
}*p = NULL, *top = NULL, *save = NULL, *ptr;
void push(int x)
{
    p = new node;
    p->data = x;
    p->next = NULL;
    if (top == NULL)
    {
        top = p;
    }
    else
    {
        save = top;
        top = p;
        p->next = save;
    }
}
char pop()
{
    if (top == NULL)
    {
        cout<<"underflow!!";
    }
    else
    {
        ptr = top;
        top = top->next;
        return(ptr->data);
        delete ptr;
    }
}
int main()
{
    std::cout << "write formula in prefix notation" << std::endl;

    tokens_t tokens;
    std::copy(
        std::istream_iterator< std::string >( std::cin ),
        std::istream_iterator< std::string >(),
        std::back_inserter( tokens ) );

    pre_to_in( &tokens );
for (int i = 0; i < strlen(x); i++)
    {
        if (x[i] >= 48 && x[i] <= 57)
            push(x[i]-'0');
        else if (x[i] >= 42 && x[i] <= 47)
        {
            a=pop();
            b=pop();
            switch(x[i])
            {
            case '+':
                push(a+b);
                break;
            case '-':
                push(a-b);
                break;
            case '*':
                push(a*b);
                break;
            case '/':     
                push(a/b);
                break;
            }
        }
    }
    cout<<"ans is "<<pop()<<endl;
    getch();
}

}