Calculator Recursion Version *HELP*

This is a discussion on Calculator Recursion Version *HELP* within the C++ Programming forums, part of the General Programming Boards category; Brief Description: User enter an expression such as "1+3+(2*5)" and the result is displayed i.e. 14. Here the code i ...

  1. #1
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    Join Date
    Apr 2006
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    3

    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.
    Last edited by _john_; 04-18-2006 at 05:44 AM.

  2. #2
    Registered User
    Join Date
    Jan 2005
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    7,328
    What's not working with your attempt to implement those last two features?

  3. #3
    Registered User
    Join Date
    Apr 2006
    Posts
    3
    I dun knw how to modify the existing code so as to take into consideration those two features... any hint will be helpful!!

  4. #4
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    Apr 2006
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    Anyone can help??

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