Thread: String Manipulation

  1. #1
    Registered User
    Join Date
    Sep 2002

    String Manipulation

    If I have a string input, something like A*B+(C+A)*B, and I do the operation in the parentesis, what would be the best way to replace what is in the parenthesis and including them to some symbol like D. For example, After I do some arithmetic on C+A, I now want the string to read:
    A*B+D*B, where D has replaced the parenthesis. Should I just have two pointers, each one pointing to a parenthesis, and dereference the second one, copy the contents into the deferenced first pointer field, and implement the second and repeat until newline? Sorry if this was unclear

  2. #2
    Join Date
    Aug 2001
    One way to do it would be to write code that
    seperates the sentence. What you want to do
    is write a function that returns the next token. So
    repeativly calling the function on the input stream
    A*B+(C+A)*B gives
    A , *, B, +, (, C, +, A, ), *, B
    You would also need to store some data in the token
    such as what value are the variables, what kind of
    token it is etc. Another function that is used
    is match. match takes token as one of the paremeters. If the current token
    equals that token it reads another token into the current
    token else it prints out an error.

    Now to do this properly you could write
    a symbol table but I think you can probably just
    do double var_list[256]. So then the value of
    C is var_list['C' - 'A'].

    The hardest part of your program is going to be
    the parser but once you get good at
    recursive descent it's really mechanical.
    There is one more thing that you need to do.
    You need to write the grammer to the language.
    I think this one will work
    prog: <expr>
    expr: {<factor> + <factor>} | {<factor> - <factor>} | <factor>
    factor: {<base> * <base> } | {<base> / <base> } | <base>
    base: <var> | ( <expr> )
    var: 'A' | 'B' .. | 'Z'

    you read this like this. prog is defined to be an expr.
    An expr is series of factor+factor or a series
    of factor - factor or just a factor.
    A factor is a series of base * base or a series of base/base
    or just a base.
    A base is just a variable or an expression grouped by
    parens. This is recursive; you must make sure
    there's no left recursion in recursive descent.
    A variable is just the uppercase letters.

    To write the recursive descent parser for this grammer
    all you need to do is to write functions prog, expr,
    factor, base, var.

    For example here is how I would start out roughly

    /* prog: <expr> */
    double prog()
          return expr();
    /* expr: {<factor> + <factor>} | {<factor> - <factor>} | <factor>
    double expr()
         double eval;
         Token next_token;
         int quit = 0;
         eval = factor();
                /* retrives the current token from the lexer */
                next_token = lookup();
                 case PLUS:
                        eval += factor();
                 case MINUS:
                        eval -= factor();
                        quit  = 1;
          return eval;
    factor is defined similarly
    /* base: <var> | ( <expr> )  */
    double base()
           Token next_token;
           double eval; 
           next_token = lookup()
           if (next_token.type == LEFT_PARAN)
                  eval = expr();
            else if (next_token.type == VAR)
                  eval = next_token.value;
                   error("Expected a '(' or variable");
            return eval;
    You will want to change some this stuft but you
    should have no problem doing this.

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