Thread: problems on polish notation

i have an algorithm on polish notation

content of the file eq.txt

Code:

((3+2)*5)+5

i will open the code using file io

Code:

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <iostream.h>
#include "stack.h"
void main()
{
clrscr();
char str[1500], file[10], sfile[10];
int x;
FILE *fp1, *fp2;
printf("Enter filename:");
gets(file);
fp1=fopen(file,"r");
printf("Save to:");
gets(sfile);
fp2=fopen(sfile,"w");

if(fgets(str, sizeof(str), fp1)!=NULL)
{
fputs(str,fp2);
printf("Content is:\n %s", str);
}
else
{
printf("cannot open file!!!");
}
fclose(fp1);
fclose(fp2);
getch();
}

the result is the equation ((3+2)*5)+5

but if this program reads the equation which is the content of the file and the output of this program will be the result of this equation

Code:

((3+2)*5)5+5
(5*5)+5
25+5
30

so parse the equation yourself or use a 3rd party maths parser to do it for you. And you haven't asked any questions, thats just what I assume you wanted.

Originally Posted by zacs7

so parse the equation yourself or use a 3rd party maths parser to do it for you. And you haven't asked any questions, thats just what I assume you wanted.

but how?

http://muparser.sourceforge.net/ Which is a C++ math parser library, but there are wrappers available, plus your "stack.h" in the other thread is in C++... So assumed your using C++ not C.

Originally Posted by zacs7

http://muparser.sourceforge.net/ Which is a C++ math parser library, but there are wrappers available, plus your "stack.h" in the other thread is in C++... So assumed your using C++ not C.

but how and where do you evaluate an expression?

Nevermind all that advanced crap, there's a whole bunch of basic horrors which need to be fixed.

> #include <stdio.h>
This is a C header

> #include <conio.h>
This is an ancient platform specific header

> #include <iostream.h>
This is an old C++ header

> void main()
Bzzzt - try again

> gets(file);
Woohoo - next stop, buffer overflow city, all change please!

Originally Posted by Adak

I STRONGLY recommend that you goto Wikipedia's website and read up on Polish Notation. You need to know EXACTLY what it is that you're trying to work with - is it:

*Polish Notation
*Reverse Polish Notation
or
Postfix Notation

You surely will want to see some examples of each of these, and decide what type of notation you want to work with, before engaging a lot of time into the project.

If you want to parse it all out, a good start would be:

1) Scan the digits and locate the innermost bracketed items.
2) Follow the rules of mathematical precedence and solve that one item.
3) Re-write the original equation, substituting your result for the innermost bracketed item.

4) Loop back to number 1 above, and repeat until it's reduced fully to a single value.

There are several sites that explain the above much better than I can. If you have trouble with your CODE or algorithm, please post your code and explain what the problem is and what output you're currently getting from your program that's not correct.

You need to be quite specific in order to get valuable help.

Good luck, this is an ambitious project.

algorithm:

Code:

*open a file
*read the file character by character
*for( i=0;i<x;i++)//assuming that x is the number of times you operate inside the parenthesis, for example, ((2+3)5)+5, this requires three times in parsing to get the result
*follow step 1 in reading characters if it is digit, place it on a stack, if it is an operand solve the number of digits inside the parenthesis
if the operand is addition or so, follow step number 2
**if it is addition, add the two numbers on a stack
**same through for subtract, multiply and divide and also solving for exponents
*the result will be the first number to be place on a stack for another parsing until the parsing is finished
*display the result

pseudo code

Code:

main()
{
in(filename)
f1=open(filename, read)
in(saved filename)
f2=open(filename, write)

if(filename!=NULL)
{
show original content
}
else
{
error
}
int x
for(i=0;i<x;i++)
{
isdigit
   placing stack
isoperand
   do step number 2
}

return result
}

but im find hard to convert my pseudo code to my C++ code, on my algorithm translate the code is my problem..how to convert my algorithm/pseudo code to C++ code?

NOTE: stack.h is include in the program

Originally Posted by Adak

If you're working in C++, then you should go to the C++ forum.

You should also read and study the several examples of what you want to do, on the net. Thats's why you can't make up good pseudo code yet.

You have a good general idea of what you want to do, and how you want to do it, but it's lacking in the detail it needs - and just what you would likely gain from - reading and studying the websites that deal with RPN or postfix, in great detail.

You need details to your pseudo code, and these websites have the details - go and get them, they're free!!

where are the website in relation to this notation?

i have an idea:

after reading the content of the file, follow the conversion of infix to postfix code below

Code:

main(int argc, char argv[])
{
	char *a=argv[1];
	int i, N=strlen(a);

	STACKinit(N);
	for(i=0;i<N;i++)
	{
		if((a[i]=='+')||(a[i]=='*'))
		    STACKpush(a[i]);
		if(a[i]++')')
		    printf("&#37;c",STACkpop());
		if((a[i]>='0')&&(a[i]<='9'))
		    printf("%c", a[i]);
	}
	printf("\n");
}

then the evaluation of postfix:

Code:

main(int argc, char argv[])
{
	char *a=argv[1];
	int i, N=strlen(a);

	STACKinit(N);
	for(i=0;i<N;i++)
        {
                if(a[i]=='+')
                   STACKpush(STACKpop()+STACKpop());
                if(a[i]=='-')
                   STACKpush(STACKpop()-STACKpop());
                if(a[i]=='*');
                   STACKpush(STACKpop()*STACKpop());
                if((a[i]>='0')&&(a[i]<=9))
                   STACKpush(0);
                while((a[i]>='0')&&(a[i]<='9'))
                   STACKpush(10*STACKpop()+a[i++]-'0'));
        }
        printf("%d \n", STACKpop());
}

how to combine these code to FILE IO on the first page?

Code:

int main ( ) {
  char *tests[] = {
    "1",
    "1+2",
    "((3+2)*5)+5",
  };
  stack s;
  double answer;
  int i;

  for ( i = 0 ; i < sizeof(tests)/sizeof(*tests) ; i++ ) {
    StackInit(s);
    parse( tests[i], s );
    answer = eval(s);
    printf("Expr=%s, Result=%f\n", tests[i], answer );
  }

  return 0;
}

No I/O to worry about, and you can add more and more complicated tests to the array as you go, and keep all the earlier tests around to make sure you don't break anything.

When you're happy with it, then you can worry about the I/O you really want.

what are the header file do you used in this program?