shunting-yard

Trouble understanding what to do with output of shunting-yard algorithm

由 自闭症网瘾萝莉.ら 提交于 2019-12-20 10:55:15
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

How can I modify my Shunting-Yard Algorithm so it accepts unary operators?

由 ぐ巨炮叔叔 提交于 2019-12-18 03:05:10
问题 I've been working on implementing the Shunting-Yard Algorithm in JavaScript for class. Here is my work so far: var userInput = prompt("Enter in a mathematical expression:"); var postFix = InfixToPostfix(userInput); var result = EvaluateExpression(postFix); document.write("Infix: " + userInput + "<br/>"); document.write("Postfix (RPN): " + postFix + "<br/>"); document.write("Result: " + result + "<br/>"); function EvaluateExpression(expression) { var tokens = expression.split(/([0-9]+|[*+-\/()

Shunting-Yard Validate Expression

由 落爺英雄遲暮 提交于 2019-12-17 20:58:24
问题 We use the Shunting-Yard algorithm to evaluate expressions. We can validate the expression by simply applying the algorithm. It fails if there are missing operands, miss-matched parenthesis, and other things. The Shunting-Yard algorithm however has a larger supported syntax than just human readable infix. For example, 1 + 2 + 1 2 1 2 + are all acceptable ways to provide '1+2' as input to the Shunting-Yard algorithm. '+ 1 2' and '1 2 +' are not valid infix, but the standard Shunting-Yard

Shunting-yard Algorithm with functions debugging

由 五迷三道 提交于 2019-12-14 02:20:06
问题 I want to implement "functions" in the shunting-yard algorithm beside operators and make a little interpreter from the resulting algorithm but syntactic incorrect usage of tokens is ignored by the default algorithm. is there anyone that has written a interpreter (or not) wanting to help me? that would help a lot of people that are stuck with this problem! here are some tests listed, the shunting-yard function ignores the wrong usage of tokens in function calls and/or missing operators

convert infix to Rpn (shunting yard)

由 て烟熏妆下的殇ゞ 提交于 2019-12-12 05:15:16
问题 here is my code to convert infix to ron using shunting yard . I know how the algorithm works well and I have no problem with that . but when I run this just nothing happen . when I debug it I get unknown error on stack initialization line #include <iostream> #include <string> #include <cstring> #include <cstdlib> #include <stack> using namespace std ; void convert (char *) ; double eval(char *) ; int precedence(char); int main() { cout << "\n Enter an expression :\n" ; cout << " >> " ; char

Modifying the Shunting Yard Algorithm (c++)

由 点点圈 提交于 2019-12-12 00:59:30
问题 I have a shunting yard algorithm in proper working order, but I noticed a special quirk: 1 + ( 3 * ( 4 + 5 ) ) parses correctly to 1 3 4 5 + * +, but 1 + (3 * (4 + 5)) fails, and parses to 1 * + 5)) + I want to get it to parse the second problem properly, so that the result is the same as the first. How can I accomplish this? Notes: I derived my algorithm from wikipedia: http://en.wikipedia.org/wiki/Shunting-yard_algorithm#The_algorithm_in_detail My algorithm code is: string switchingYard

Abstract syntax tree using the shunting yard algorithm

由 放肆的年华 提交于 2019-12-10 16:56:54
问题 I have an infix expression that I have tokenised and would like to proceed to create an abstract syntax tree. I understand the shunting-yard algorithm used in these cases. I have only found ways to convert the infix expression to RPN format, not into an AST. I could create the RPN version first and then AST from it, but it seems unnecessary. My language of choice is JavaScript, though I only need to see an example in any language and/or a description of the algorithm. I have skimmed through

Infix to postfix algorithm that takes care of unary operators

由 不羁的心 提交于 2019-12-09 18:18:41
问题 The I/p to the algo will be an expression like this: a+(-b) a*-b+c i.e any expression that a standard C compiler would support. Now I've the input already formatted as a stream of tokens , the tokens contain info whether its an operator or an operand. The algorithm should take this in and give me a postfix expression that I can evaluate. If I use the standard conversion algo, I cant differentiate between an unary and a binary op . Like a*(-b) would give me ab-* ,which would evaluate in the

Postfix to infix with unary/binary operators

由 梦想的初衷 提交于 2019-12-08 06:29:58
问题 I am trying to make a converter from postfix to infix notation and need some help. There is already a question about infix-to-postfix conversion, which gives an example I am failing to convert back. (Note: a minus sign is missing there!) The following is the output of my converter, where the 1st "column" is postfix input, the 2nd is my infix output, and the 3rd is what I probably should get(?): 2 - = - 2 =? - 2 true 1 + 2 + = + 1 + 2 =? + 1 + 2 true 1 + 2 + + = + (+ 1 + 2) =? + 1 + + 2 false

Shunting-yard algorithm in c++

由 ﹥＞﹥吖頭↗ 提交于 2019-12-06 01:48:57
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 = "0123456789"; string posops = "+-*/%(){}[]"; string onlyops = "+-/%*"; string space = " "; string openbra =
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