Thread: reverse polish calculator

I took on these exercises related to the reverse polish calculator mentioned in K&R book.

And there is this exercise which is giving me a headache:

Add commands for handling variables. (It's easy to provide twenty-six variables with single-letter names.) Add a variable for the most recently printed value.

How exactly am i supposed to do this and what am i supposed to do? Does this mean enabling = operator for giving values to single letter variables? If so, how to implement that sort of design... i mean it has to work with the stack somehow, im just confused how... and how many variables could be at use in one time, all 26 or just one?

The code for the calculator is below:


(seperated into 5 source files with appropriate functions)

Code:

#include "calc.h"

#define MAXOP 100
int main(int argc, char *argv[])
{
    
    int type;
    double op2;
    char s[MAXOP];
    
    while((type = getop(s)) != EOF)
    {
       switch(type) {
         case NUMBER:
              push(atof(s));
              break;
         case WORD:
              Handle_word(s);
              break;
         case '+':
              push(pop() + pop());
              break;
         case '*':
              push(pop() * pop());
              break;
         case '-':
              op2 = pop();
              push(pop() - op2);
              break;
         case '/':
              if((op2 = pop()) != 0)
                push(pop() / op2);
              else 
                printf("zero divisor\n");
              break;
         case '%':
              if((op2 = pop()) != 0)
                push(fmod(pop(), op2));
              else
                printf("zero divisor\n");
              break;
         case '\n':
              printf("\t%g\n", pop());
              break;
         default:
              printf("unknown command: %s\n", s);
              break;
       }
    }
    
     getchar();
     return 0;
}

Code:

#include "calc.h"
#define MAXVAL 100

double val[MAXVAL];
int sp = 0;

void push(double c)
{
     if(sp < MAXVAL)
       val[sp++] = c;
     else
       printf("stack full\n");
}

double pop()
{
     if(sp > 0)
       return val[--sp];
     else {
       printf("stack empty\n");
       return 0.0;
     }
} 

void Handle_word(char s[])
{
    double op1, op2;
    
    if(strcmp(s, "print") == 0)
    {
      if(sp > 0)
        printf("\t%g\n", val[sp-1]);
      else 
        printf("stack empty, cant print\n"); 
    }
    else if(strcmp(s, "dup") == 0)
    {
      op1 = pop();
      push(op1);
      push(op1);
    }
    else if(strcmp(s, "swap") == 0)
    {
       op1 = pop();
       op2 = pop();
       push(op1);
       push(op2);
    }
    else if(strcmp(s, "clear") == 0)
    {
       sp = 0;
    }
    else if(strcmp(s, "sin") == 0)
    {
       push(sin(pop()));
    }
    else if(strcmp(s, "exp") == 0)
    {
       push(exp(pop()));
    }
    else if(strcmp(s, "pow") == 0)
    {
       op1 = pop();
       push(pow(pop(), op1));
    }
    else 
      printf("unknown command: %s\n", s);
}

Code:

#include <ctype.h> /* isdigit */
#include "calc.h"

/* get a operand or operator, or a word */
int getop(char s[])
{
    int i = 0, c, next;
    
    while((s[0] = c = getch()) == ' ' || c == '\t') /* the first character */
          ;
    s[1] = '\0'; 
    
    
    /*1. if the first char is a letter, handle the string as a word */
    if(isalpha(c)) 
    {
       while(isalpha(s[++i] = c = getch()))
            ;
       
       ungetch(c);
       s[i] = '\0';
       return WORD;
    }
    
    /*2. Handle numbers, possible unary minus - if not any of that just return */
    if(!isdigit(c) && c != '.' && c != '-')
      return c;
    
    /* from now on, c can only be . or digit or - */
    if(c == '-')
    {
         next = getch();
         
         if(!isdigit(next) && next != '.') { /* . because -.25 is allowed */
           ungetch(next);
           return c; /* not a number, but an operator, so we must return it! */
         }
         
         c = next; /* if next is a . or a digit, just put it into c */
    }
    else /* if c is a . or a number, get another char since . or number is allready stored in */
       c = getch();
    
    /* now c can only be . or - */
    while(isdigit(s[++i] = c))
       c = getch();
    
    if(c == '.')
      while(isdigit(s[++i] = c = getch()))
           ;
    
    s[i] = '\0';
    if(c!=EOF)
      ungetch(c);
    
    return NUMBER;
}

Code:

#include "calc.h"
#define MAXBUF 100

int buf[MAXBUF];
int bufp = 0;

int getch()
{
    return (bufp > 0) ? buf[--bufp] : getchar();
}

void ungetch(int c) //stores the temp char
{
     if(bufp >= MAXBUF)
        printf("ungetch: too many characters\n");
     else
        buf[bufp++] = c;
}

Code:

/* headers */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

#define NUMBER '0' /* used by main and getop */
#define WORD 'a'

int getop(char []);

void push(double);
double pop();

int getch();
void ungetch(int);

void Handle_word(char []);

Originally Posted by Tool

Add commands for handling variables. (It's easy to provide twenty-six variables with single-letter names.) Add a variable for the most recently printed value.

How exactly am i supposed to do this and what am i supposed to do?

Define 26 alphabet variables.

Originally Posted by Tool

Does this mean enabling = operator for giving values to single letter variables?

Yep!

Originally Posted by Tool

If so, how to implement that sort of design...

Simply extend your case statement so it can parse the equality operator and the 26 alphabetic variables.

Originally Posted by Tool

i mean it has to work with the stack somehow, im just confused how...

Need to push() the variables onto the stack exactly like you have done before, and when you see the equals sign you pop() it and assign the value to the variable.

Originally Posted by Tool

and how many variables could be at use in one time, all 26 or just one?

You will allocate 26 variables but use only the subset that has been entered by the user at runtime.