postfix-notation

问题 I want to create an expression tree given expression in infix form. Is it necessary to convert the expression to postfix first and then create the tree? I understand that it somehow depends on the problem itself. But assume that it is simple expression of mathematical function with unknowns and operators like: / * ^ + -. 回答1: No. If you're going to build an expression tree, then it's not necessary to convert the expression to postfix first. It will be simpler just to build the expression tree

问题 I'm currently working on a project to convert from postfix to infix using a stack in the form of a singly linked list. I've managed to convert expressions such as ab+ to (a+b) however when the expression gets longer such as ab+cd*- . It doesn't work. I'm considering pushing the previously converted expression back onto the stack however the stack is of type char and the expression is a string and it complains when I try to push it back. Should I make it a template and if so how would I do it

问题 I'm trying to write a program that evaluates a postfix arithmetic expression. The program sends a character string to my function evaluatePostfix , which proceeds to identify operands and operators and come up with an integer solution. I am manipulating stacks in this program by pushing the scanned character as it is identified and of course doing the appropriate pop functions when needing to evaluate. Right now though, I'm having a problem with the program hanging in what appears to be an

问题 I'm writing a program that takes user input and uses stacks to convert an infix expression into a postfix expression based on precedence, with operands always going before operators. For example, if a user inputs: (a+b*c) then the program should display: abc*+ so far, I have this: #include <iostream> #include <stack> #include <string> using namespace std; int main() { stack<char> s; char input; while (cin.get(input) && input != '\n') { if (isalnum(input)) cout << input << "\n"; else if (input

问题 i recently wrote a java program that takes an infix expression and converts it into a postfix expression. It works for the most part but i am getting wrong outputs for some expressions. For example the expression a+b+c+d+e will output abcde+++++ when it should output a b + c + d + e +. import java.util.Stack; public class ITP { public static Stack<Character> stack; public static String inFixExp; public static String postFixExp = ""; public static String infixToPostfix(String exp){ ITP o = new

问题 I'm trying to understand a topic in class about using stacks and queues as a means of programming a calculator. I understand what infix and postfix expression is but how does it make it easier for a program to evaluate an expression and why are queues and stacks ideal in this situation? Thanks 回答1: It makes the order of operations simpler to handle, for example: + * - 4 2 5 3 Can only mean ((4 - 2) * 5) + 3 Which might be more readable for us, but we need to know the order of operations and

问题 I have been working on this infix to postfix/polis notation converter. Although, I do not feel the solution is adequate. Specifically the j (EDIT: Now called index) variable is bugging me. Do you guys have any suggestions? Or perhaps there is a much better way to accomplish it? Or do I just worry too much? public static string[] InfixToPostfix(string[] infixArray) { var stack = new Stack<string>(); var postfix = new string[infixArray.Length]; int index = 0; string st; for (int i = 0; i <

问题 I need a function that takes an infix string (like "3 + 4 * 9"), and convert it to postfix (like "4 9 * 3 +"). I got it working until you throw in parentheses within parentheses. I've been working on it all day and can't figure out what I'm doing wrong- can someone with a fresh mind see it, maybe? I feel like I'm really close! Thanks! Here's the code: string ExpressionManager::infixToPostfix(string infixExpression) { cout << "itop Testing : " << infixExpression << endl; string posnums =

问题 I've been mulling over creating a language that would be extremely well suited to creation of DSLs, by allowing definitions of functions that are infix, postfix, prefix, or even consist of multiple words. For example, you could define an infix multiplication operator as follows (where multiply(X,Y) is already defined): a * b => multiply(a,b) Or a postfix "squared" operator: a squared => a * a Or a C or Java-style ternary operator, which involves two keywords interspersed with variables: a ? b

问题 I've been looking at the wiki page: http://en.wikipedia.org/wiki/Shunting-yard_algorithm I've used the code example to build the first part, basically I can currently turn : 3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3 into 3 4 2 * 1 5 − 2 3 ^ ^ / + But I don't know how to then use 3 4 2 * 1 5 − 2 3 ^ ^ / + to obtain 3.00012207 And the example code and explanation on wiki aren't making any sense to me. Could someone please explain how to evaluate 3 4 2 * 1 5 − 2 3 ^ ^ / + and produce the answer. Thanks in