postfix-notation

问题 I searched all over the internet for a good implementation in converting not numbers expressions but variable expressions from infix notation into prefix and postfix. All the searches I did weren't successful. Basically I want to see if there is any implementation yet in PHP so I could modify it to support more operators not only the (-,*,+,=). For example convert: a+b/c*(p/c) While keeping the variable names, and not having to enter numbers to evaluate them. 回答1: I have found a good

问题 There are many algorithms to convert infix to postfix all over the web. But my question is how to make that to support functions? For example sin(x+y)*z. I will appreciate a code. 回答1: If you are looking for an algorithm that gives you the conversion infix to postfix including function call support, you can use the below pseudocode(which looks like python code). I have written this for my case but not yet tested thouroughly. If you find any bugs please let me know. I have also written a Java

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 7 years ago . I can not solve this expression from postfix to infix. Please help me to understand in detail 5 x y - / x y + 3 ^ 7 / + 回答1: postfix to infix: 5 x y - / x y + 3 ^ 7 / + STEP 5 x y - / A) 5xy-/ = 5 (x-y)/ = (5 /

问题 I'm working on a project in Java that requires me to convert an infix expression to a postfix expression. I am currently able to convert infix expressions to postfix with this method as long as they don't contain parenthesis, but I can't figure out how to handle parenthesis. Basically, I have two stacks that hold objects that are called 'Token'. A Token is a wrapper class that holds a string that is either a number, variable (which gets evaluated as a number, pending on user input), operator

问题 I'm working on a project in Java that requires me to convert an infix expression to a postfix expression. I am currently able to convert infix expressions to postfix with this method as long as they don't contain parenthesis, but I can't figure out how to handle parenthesis. Basically, I have two stacks that hold objects that are called 'Token'. A Token is a wrapper class that holds a string that is either a number, variable (which gets evaluated as a number, pending on user input), operator

问题 I read book today. It introduced algorithm about chaning from infix to postfix.. What is benefits? Thanks in advance. 回答1: Well for one, you can easily evaluate a postfix expression in a single scan from left to right with the help of a stack, unlike evaluating infix expressions. and second, there is no need of the concept of parentheses and precedence rules etc. in a postfix expression. 回答2: I think infix is really easy to understand for human. Postfix is the good way for machine to process.

问题 I'm writing what might not even be called a language in python. I currently have several operators: + , - , * , ^ , fac , @ , !! . fac computes a factorial, @ returns the value of a variable, !! sets a variable. The code is below. How would I go about writing a way to define functions in this simple language? EDIT: i updated the code! import sys, shlex, readline, os, string List, assign, call, add, sub, div, Pow, mul, mod, fac, duf, read,\ kill, clr, STO, RET, fib, curs = {}, "set", "get", "+

问题 I am trapped at work with a locked down pc. But I am trying to practice my Scala. I am using Ideone.com since I can't even install scalac ... Anyway this is not compiling: class DPt(var name: String, var x: Double, var y: Double){ def print = println("DPt; name: " + name + " x: " + x + " y: " + y) } object Main { def main(args: Array[String]) { val pt0 = new DPt("Joe", 1.0, 1.0) println("checking generated accessor: " + pt0.x) pt0 print pt0.x_=(3.0) pt0 print } } I get this message back from

问题 Good day! I am implementing an infix to postfix converter using stacks. It works when the user input an infix expression with no parenthesis; but when a parenthesis is present, the console says: Exception in thread "main" StackEmptyException: Stack is empty. at ArrayStack.top(ArrayStack.java:85) at InfixToPostfix.convert(InfixToPostfix.java:54) at InfixToPostfix.main(InfixToPostfix.java:85) My problem is in implementing the rank (top of the stack). 回答1: Aha! You need "stack peek" when

问题 I was given an assignment to write a program that evaluates a postfix expression using the stack. I wrote the program and it seems to be working for the most part, however it is having issues determining whether the expression is valid. Here are the basic steps that I've done: Ask user to input an expression (store as string) Iterate through each char and determine if the char is an operand, operator, space, or invalid character if(char == operand) push onto stack if(char == operator) pop