convert infix to postfix expression and show trace of algorithm?
Q. Convert the infix expression (A - B) * C + D to postfix. Show the trace of the algorithm, i.e., the stack, the infix expression and postfix expression.
Asked by bc050400307 - Wed Aug 15 00:29:21 2007 - - 1 Answers - 0 Comments

A. a b - c * d + stack trace. [a-b] means a single element that represents the value of a-b. a a b a b - [a-b] [a-b] c [a-b] c * [(a-b)*c] [(a-b)*c] d [(a-b)*c] d + [(a-b)*c + d]
Answered by anonymous - Wed Aug 15 01:12:10 2007

Please tell me how do you convert infix expression to prefix expression in C++ programming?
Q. With and without using stacks.
Asked by RJ - Thu Mar 15 15:18:38 2007 - - 1 Answers - 0 Comments

A. The Shunting Yard Algorithm does this using a stack. I'm not sure how to do it without a stack.
Answered by Bill Lumbergh - Fri Mar 16 03:22:24 2007

Can you give me a program to convert infix expression to a postfix expression?
Q. the program should be in c language and implement stack in it. please give me a working code or a link to the same
Asked by anku - Mon Aug 31 01:58:29 2009 - - 2 Answers - 0 Comments

A. You can read the expression put each element it in the stack and retrieve it and place the operators after the operands. Here is a link that answers your question : Don't copy the code from there just use it as a reference else it will be cheating and you will not learn anything. Good Luck!!!
Answered by abc - Mon Aug 31 02:31:31 2009

Write a program to convert the infix input expression into postfix expression?
Q. Example Input infix expression 5*(((9+8)*(4*6) +7) Result postfix expression 598+46**7+*
Asked by swetha b - Sun Aug 19 01:22:26 2007 - - 1 Answers - 0 Comments

A. Do your own homework.
Answered by Popsmear - Sun Aug 19 01:29:07 2007

write a program to construct a binary tree using giver prefix and infix expression?
Q. write a program to construct a binary tree using giver prefix and infix expression?
Asked by pooja s - Thu Dec 11 22:41:18 2008 - - 1 Answers - 0 Comments

A. Enjoy.
Answered by feynman_rocks - Sun Dec 14 22:00:29 2008

How to convert an infix expression into a prefix expression using stack class in C++.?
Q. How to convert an infix expression into a prefix expression using stack class in C++.?
Asked by ADS - Thu Nov 2 16:42:51 2006 - - 2 Answers - 0 Comments

A. 1.21 Gigawatts is way too much. You'll destabilize the dilithium crystals. This is an interesting question, most likely a well thought out assignment for a class. So, I think you'll benefit from doing it yourself. I just wrote you a big long answer that suggested a solution, but I'm deleting it, because I think it might be right. Instead, I'll give you a hint: Can you use more than one stack? And should we apply operands to the top two values in a stack of numbers? Hold on, you want to convert to prefix, not evaluate. Okay, but I think what I have in mind still holds. Also, you might think about drawing a parse tree for the expression. It might help you to see what's going on.
Answered by arbeit - Thu Nov 2 20:21:08 2006

Vectors and Evaluating Infix Expression C++?
Q. I've been racking my brain on this all day and can't seem to figure it out. I have to vectors. One has int of numbers and the other has string of ops. I have to go through the vectors and evaluate for the result. So if the infix expression is 5 + 3 * 2 - 8 / 2 I would get a vector containing: 5 3 2 8 2 and another vector containing: + * - / Now how do I go through the vector of op and find the first * or / or % and do the operation of the right two numbers in the vector of int and then insert the result into the right position in the vector of int and then remove the op from the vector of op and keep doing this for the other * / or %. Then go through the op vector in order and add or subtract to the first two numbers in the vector and… [cont.]
Asked by justinsbabe5000 - Wed Jun 3 00:21:33 2009 - - 3 Answers - 1 Comments

A. The key is your vector of ops.Since you are using only +-*/, you just start through the ops vector and first look for the * and / characters. Each time you find one, you use its vector index to access the surrounding two numbers in the int vector. Using ASCII character matching, determine which operation to perform, update one of the int vectors with the new value and eliminate the other, then eliminate the op vector. Once all the * and / ops are gone, do the same for the + and - ops. This code is a little too complex to simply post, hopefully you have a working vector with update and remove functions, and know how to loop through the vectors and keep track of position.
Answered by HandyManOrNot - Wed Jun 3 00:58:42 2009

How to parenthesize an infix expression?
Q. here'e my code: string p_exp; int k=0; p_exp[1]='('; for(int j=2;j<=e.length()+1;j++) //"e" is the infix expression { k++; p_exp[j]=e[k]; } p_exp[j]=')'; But there's no output, i dont know wats wrong with the code. Could anyone kindly help me n suggest me if there's some other way to parenthesize this expression.I wish to place parenthesis around the whole expression for e.g. (x+2=10) Do reply me soon Thankyou!
Asked by Hazel - Tue Oct 16 09:51:08 2007 - - 1 Answers - 0 Comments

How can I implement infix to prefix expression in C language?
Q. How can I implement infix to prefix expression in C language?
Asked by Raju Kumar S - Sun Apr 1 16:47:16 2007 - - 2 Answers - 0 Comments

A. You may have already gotten this in the class you are doing this for, but you should build and use a tree.
Answered by wassabison - Sun Apr 1 22:29:58 2007

how to convert infix expression to postfix form?
Q. how to convert infix expression to postfix form?
Asked by ann_svit - Wed Aug 22 18:37:37 2007 - - 2 Answers - 0 Comments

A. In postfix, the operators are written after the operands. For example, if you have an infix expression.. a + b The postfix would be a b + A more complicated example (infix first).. (z + y) * a - c The postfix would be z y + a * c - If you think of an expression being put on a stack and popped whenever an operator is encountered it might be easier. For instance.. z and y would be placed on the stack. The + is encountered and both z and y are popped and added together. The result is put back on the stack. The a is the put on the stack and then the * is encountered. Both the a and the result of z + y are popped and multiplied. The answer is put back on the stack. Then the c is placed on the stack. Once the minus is encountered, [cont.]
Answered by marleymang - Wed Aug 22 21:50:38 2007

write a stack program that inputs infix expression the outputs postfix expression?
Q. for example (6+9-2)*5/2, its output is 69+2-5*2/
Asked by litol_twilky - Thu Aug 9 20:40:25 2007 - - 1 Answers - 0 Comments

A. It's easy to do this with a stack -- just match parenthesis and pust to the stack based on order of ops
Answered by mdigitale - Thu Aug 9 20:43:03 2007

can you give me a c++ program that accepts postfix or prefix expression and convert it to infix as output?
Q. can you give me a c++ program that accepts postfix or prefix expression and convert it to infix as output?
Asked by ugat_puno - Fri Jul 27 11:49:22 2007 - - 3 Answers - 0 Comments

Java infix to postfix question?
Q. i got most of it. and i cant figure out one thing that is operating in in fix and doing the arithmatic import java.util.Stack; import java.util.Queue; import java.util.LinkedList; import java.util.Scanner; /** * This program lets you type in a fully parenthesized infix expression, then * it converts the expression to postfix, and evaluates the postfix expression. * Only works with single-digit integers and +, -, /, *. * It displays an error message for division by zero. * Spaces are allowed anywhere in the expression. * * @author (your name) * @created October 29, 2008 */ public class InfixToPostfix { /** * The main program for the InfixToPostfix class. * * @param args The command line arguments. */ public static void main(String[] args)… [cont.]
Asked by twitch - Sun Nov 2 16:54:11 2008 - - 2 Answers - 0 Comments

A. import javax.swing.JOptionPane; public class InToPost { private LinkList list; private String inString; private String outString = ""; public InToPost myStack; public InToPost(String in) { inString = in; myStack = new InToPost(); } public InToPost() { list = new LinkList(); } public void push(char j) { list.insertFront(j); } public char pop() { return list.deleteFront(); } public boolean isEmpty() { return (list.isEmpty()); } public void displayStack() { System.out.print("Stack: "); list.displayList(); } class LinkList { private Link head; public LinkList() { head = null; } public boolean isEmpty() { return (head == null); } public void insertFront(char d) { Link newLink public char deleteFront() { Link temp = head; head =… [cont.]
Answered by astrid - Mon Nov 10 08:00:49 2008

How do you get a file name from a user in C?
Q. I have an assignment where I need to request a user for a file name and then read the characters from the file into a character array for converting an infix expression to postfix. However, I am not well versed in C's input/output syntax. Any help would be appreciated
Asked by Shelly - Sat Oct 24 20:03:26 2009 - - 1 Answers - 0 Comments

In JAVA I want to create a Postfix and Prefix from an infix, but how do I tell JAVA to compare operators?
Q. I am trying to take an infix expression such as a + b + c and turn it to postfix ab+c+ and prefix ++abc. I need to push them on the stack etc. While I worked out a method for stack pushing It depends on whether or not the operator is of a higher presidence or not than that one already on the top of the stack. The problem is that I don't know how to in JAVA compare operators. I did find some stuff such as this: StackB x = new StackB (); String op = "+-*/%^"; String post = ""; I wonder about the line: String op = "+-*/%^"; Are they somehow using this so that the presidence is tested from left to right? Where they somehow make it so that + and - are equal and *, /, and % are equal? so for example since * is more to the right than + it has… [cont.]
Asked by Brian D - Sun Nov 11 22:57:28 2007 - - 1 Answers - 0 Comments

A. Operator precedence are already defined in java ... surf through the link below ,this may help you go to the page no 96-97-98 of this book try to store all d operators in an array rather then a string ...
Answered by sameer azazi - Sun Nov 11 23:10:12 2007

Algorithm for making an expression tree for mathematical functions?
Q. Hi there, Could someone describe to me the method/algorithm for taking an infix mathematical expression like: (8 (4 2) )* ( 4+(8 + 6) ) (assume this is input as a string) And converting it into an expression tree? I cannot just use existing methods like python eval(), I need to implement this algorithm and it's driving me up the wall trying to figure out the logic ;) I know that every parenthesis means another node deeper into the tree, I also know that the root nodes should all be operators, and only the leaves operands. If it helps, I am coding this in Python with a binary tree class which I've already defined. If there are any particular methods the binary tree class requires for this (ie "new left node/right node(root node):")… [cont.]
Asked by CrArC - Mon Mar 9 12:04:04 2009 - - 1 Answers - 0 Comments

A. First off, it looks like your expression is fully parenthesized. If that's the case, we don't have to worry about operator precedence (the old "Please Excuse My Dear Aunt Sally"). Also, I am assuming there's no whitespace. I'll let you figure out how to scan over whitespace. To build an infix expression tree, we will build the tree in-order using a recursive algorithm. Let's say the expression is stored in an array called Expr, which has length Expr.length and can be queried using Expr[0]...Expr[length-1] (as in Java). For simplicity, the expression Expr[x ... y] will return a substring starting at index x and ending at index y (inclusive). We'll also define a datatype for a tree. You can either have a Number(n) or a Node(left, op,… [cont.]
Answered by MH - Thu Mar 12 14:43:38 2009

i need help in C++ data structure ?
Q. who can give me the code for this program or can give me some hints and help me with the idea to write it ? Write a driver to test your class templates and algorithms. If the infix arithmetic expression (66 + 38 - 93) * 5 / 6 - 88 % 6 is given, the postfix version of the preceding infix expression 66 38 + 93 - 5 6 */ 88 6 % - will be created and then evaluated. Thus the result is = 5
Asked by Alex Sexy - Sun Aug 20 02:58:35 2006 - - 3 Answers - 0 Comments

A. What exactly do you need ? 1) Program that converts infix expressions to prefix expressions ? 2) A Driver program that tests an infix prefix converter ?
Answered by gjmb1960 - Sun Aug 20 03:39:49 2006

Please Solve this question :?
Q. Write an algorithm in C to convert an infix expression to a postfix expression. Execute your algorithm with the following infix expression as your input. (m+n)*(k+p)/(g/h) ^| (a^| b/c
Asked by abhishek_vis - Thu Oct 18 06:57:30 2007 - - 2 Answers - 0 Comments

A. this really isn't the place to get your homework done...
Answered by UltraMagnus - Thu Oct 18 07:07:34 2007

please explain how this program works line by line?
Q. include #include #include #include typedef struct my_struct{ char data; struct my_struct *link; }item, *itemptr; char popData; void display(itemptr bottom){ itemptr p; if (bottom==NULL){ puts(" none"); } else { p = bottom; while(p!=NULL){ printf("%c", p->data); p=p->link; } } } itemptr pop(itemptr bottom, int Nodes){ itemptr p, next; int ctr=1; if (bottom == NULL){ popData = '\0'; } else { if (Nodes!=1){ p = bottom; next = bottom; do{ ctr++; next = p; p=p->link; } while (ctr!=Nodes); next -> link = p->link; popData = p->data; free(p); } else { p = bottom; popData = p -> data; free(p); } } return bottom; } itemptr push(char x, itemptr bottom, int Nodes){ itemptr Nn, p; int ctr=0; Nn = (item [cont.]
Asked by pink angel - Thu Aug 28 03:24:22 2008 - - 2 Answers - 0 Comments

A. It is a link list program, you are supposed to do your homework yourself.
Answered by Rina S - Thu Aug 28 03:32:08 2008

Why computers prefer and use Postfix expressions instead of prefix expressions??
Q. We all know computers convert all infix arihmetic expressions to postfix and then evaluate them.. but why we use and prefer postfix over prefix expressions?
Asked by Jublee - Tue Sep 26 13:42:21 2006 - - 3 Answers - 0 Comments

A. i think implementation of code for sloving postfix expression is easier and solving postfix expressions is much more faster & efficient then solving prefix expressions.
Answered by Digitally ! - Wed Sep 27 14:05:36 2006