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Help with parse tree calculator

hartman498 (4)
Hi, so i'm working on this project and i seem to keep getting a number off what it should be. I believe my problem lies in the eval function but i'm not sure. The program compiles and runs but the output is off. Any help would be great. Thanks

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double eval(Tree *root)
{
    // for (recursively) evaluating a tree; this function will refer
    // back to itself (for evaluating subtrees) and will use the
    // evalOp() functions
    
    double result;
    if(root->op != "")
    {
        if(root->right == NULL) result = evalOp(root->op, eval(root->left));
        else result = evalOp(root->op, eval(root->left), eval(root->right));
    }
    return result;
}
AbstractionAnon (6954)
Not enough information.

You haven't shown us the declaration for Tree, nor evalOp (2 argument and 3 argument versions).

From what I see, there is an implicit assumption that root->left is never null.
If root->left can be null, then the above code is going to have problems.

Last edited on
hartman498 (4)
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#include <iostream>
#include <string>
#include <cmath>
#include "math_tree.h"
#include "calc_parser.h"
using namespace std;

double evalOp(string op, double val)
{
    // for evaluating functions like sin, cos, etc.
    double result;
    if(op == "sin") result = sin(val);
    if(op == "cos") result = cos(val);
    if(op == "tan") result = tan(val);
    if(op == "log") result = log(val);
    if(op == "abs") result = abs(val);
    return result;
}

double evalOp(string op, double val1, double val2)
{
    // for evaluating operators like +, -, etc.
    double result = 0;
    if(op == "+") result = val1+val2;
    if(op == "-") result = val1-val2;
    if(op == "/") result = val1/val2;
    if(op == "*") result = val1*val2;
    if(op == "^") result = pow(val1, val2);
    return result;
}

double eval(Tree *root)
{
    // for (recursively) evaluating a tree; this function will refer
    // back to itself (for evaluating subtrees) and will use the
    // evalOp() functions
     double result;
    if(root->op != "")
    {
        if(root->right == NULL) result = evalOp(root->op, eval(root->left));
        else result = evalOp(root->op, eval(root->left), eval(root->right));
    }
    return result;
}

// this is a global variable used by the parser
Tree* math_tree;

int main()
{
    // hand-code a tree for this expression:
    //  sin(3.14159/4.0)+cos(8.0*2.0)
    Tree p1;
    p1.op = "+";
    Tree p2;
    p2.op = "sin";
    Tree p3;
    p3.op = "cos";
    Tree p4;
    p4.op = "/";
    Tree p5;
    p5.op = "*";
    Tree v1;
    v1.val = 3.145159;
    Tree v2;
    v2.val = 4.0;
    Tree v3;
    v3.val = 8.0;
    Tree v4;
    v4.val = 2.0;
    p1.left = &p2;
    p1.right = &p3;
    p2.left = &p4;
    p2.right = NULL;
    p3.left = &p5;
    p3.right = NULL;
    p5.left = &v3;
    p5.right = &v4;
    p4.left = &v1;
    p4.right = &v2;
    v1.left = v1.right = v2.left = v2.right = v3.left = v3.right = v4.left = v4.right = NULL;
    // here we print your hand-coded tree
    print_ascii_tree(&p1);

    // and evaluate the hand-coded tree (should equal -0.25)
    cout << "Result: " << eval(&p1) << endl << endl << endl;

    // this is the user-input loop; there is no need to change it
    while(true)
    {
        Parser parser;
        cout << "Enter expression: ";

        // this function gets the input and does the parsing
        parser.yyparse();

        // the yyparse() function sets the global variable math_tree
        // to a new tree; if that tree is NULL (no tree), just quit
        if(math_tree == NULL)
            break;

        // otherwise, print the tree
        print_ascii_tree(math_tree);

        // and evaluate it
        cout << "Result: " << eval(math_tree) << endl << endl;

        // reset the tree back to NULL before looping
        math_tree = NULL;
    }

    return 0;
}


this is the code so far. thanks
AbstractionAnon (6954)
Line 81: You set v1-v4 left and right to null. That makes sense, but as I pointed out earlier, if left is null in any node that is going to cause a problem.

Line 40-41: Consider what happens when when root->left is null. You call eval() with a null pointer. eval() contains no check for a null pointer. It just assumes it was passed a valid pointer.
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