Thread: Calculator Recursion Version *HELP*

Brief Description: User enter an expression such as "1+3+(2*5)" and the result is displayed i.e. 14.

Here the code i managed to get so far with the help of a friend:

Code:

#include<iostream>
#include<sstream>
#include<string>
#include<cctype>
#include<cmath>
using namespace std;

enum {PLUS='+',MINUS='-',MUL='*',DIV='/'};


double Value_Of_Num(string &Expr)


{ 

    istringstream is(Expr);
    double value=0.0;
    is >> value;
    return value;
}

double Value_Of_Expr(string &Expr)

{ 
	int i=0,p=0,pos_mul=0,pos_div=0;

    if(Expr.at(0)=='('&&Expr.at(Expr.length()-1)==')')
	{ 
		for(;i<Expr.length();i++)


            { 
			
			if(Expr.at(i)=='(') p++;
            else if(Expr.at(i)==')') p--;
            if(p==0) break; 
			
			} 

        if(i==Expr.length()-1)
        return Value_Of_Expr(Expr.substr(1,Expr.length()-2));
    }

    for(;i<Expr.length();i++)


        { 
		
		if(Expr.at(i)=='(') p++;
        else if(Expr.at(i)==')') p--;
        else if(p==0&&ispunct(Expr.at(i)))
        	
			{ 
			
			switch(Expr.at(i))


				{ 
					case PLUS:
					return Value_Of_Expr(Expr.substr(0,i))+Value_Of_Expr(Expr.substr(i+1,Expr.length()-i-1));
					case MINUS: 
					return Value_Of_Expr(Expr.substr(0,i))-Value_Of_Expr(Expr.substr(i+1,Expr.length()-i-1)); 
					case MUL: pos_mul=i; break;
					case DIV: pos_div=i; break;
					
				}

			} 

		}

	if(pos_mul)
	return Value_Of_Expr(Expr.substr(0,pos_mul))*Value_Of_Expr(Expr.substr(pos_mul+1,Expr.length()-pos_mul-1));
	
	if(pos_div)
	return Value_Of_Expr(Expr.substr(0,pos_div))/Value_Of_Expr(Expr.substr(pos_div+1,Expr.length()-pos_div-1));
	
	return Value_Of_Num(Expr);
}


int main()

{ 
string expression;

cout <<"Enter your expression:";
cin >> expression;

cout << Value_Of_Expr(expression)<<endl;
return 0;
}

It is for a homework and it's need to be of recursive type.
The code above work @ 100% but I need to implement the following additional features:
> Able to compute negative numbers
> Multiple brackets such as "((1+3) +(9/3))*2)

PLZ PLZ HELP - Any suggestions or corrections in the code above will be greatly appreciated.