infix-notation

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 am trying to remove white space that exists in a String input . My ultimate goal is to create an infix evaluator, but I am having issues with parsing the input expression. It seems to me that the easy solution to this is using a Regular Expression function, namely Regex.Replace(...) Here's what I have so far.. infixExp = Regex.Replace(infixExp, "\\s+", string.Empty); string[] substrings = Regex.Split(infixExp, "(\\()|(\\))|(-)|(\\+)|(\\*)|(/)"); Assuming the user inputs the infix expression

问题 I enjoy common lisp, but sometimes it is really painful to input simple math expressions like a(8b^2+1)+4bc(4b^2+1) (Sure I can convert this, but it is kind of slow, I write (+ () ()) first, and then in each bracket I put (* () ())...) I'm wondering if anyone here knows a better way to input this. I was thinking about writing a math macro, where (math “a(8b^2+1)+4bc(4b^2+1)”) expands to (+ (* a (1+ (* 8 b b))) (* 4 b c (1+ (* 4 b b)))) but parsing is a problem for variables whose names are

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

问题 thanks for reading. I really need some help here. I'm faced with a problem that involves me using a recursive method to store a user-input fully marked prefix expression. In this particular instance, numbers are identified with the symbol "n", while symbols and such are just as they are. So, for example, this: % 18 + 14 / ( 12 – 11 ) – 1183 % 17 Would be input into the command prompt as this: % n18 / n14 - n12 n11 % n1183 n17 Now, here's what I have for the processing of a completely

问题 I'm trying to create a binary tree like this: link The input will be entered by the user as prefix and read as a string, then put in a binary tree. This is what I have so far: struct node{ char val; struct node *left; struct node *right; }; typedef struct node root; typedef root *tree; In main: void main(){ int i; tree tr; char* s; s=input(); //input function tr=create_empty_tree(); for(i=0;s[i]!='\0';i++){ tr=add_root(s[i],ab); } convert_infix(tr); } and this is the part I've been struggling

问题 (defun solve (L) (cond ((null L) nil) (t(eval (list (car (cdr L)) (car L) (car (cdr (cdr L)))))))) The code I have is a simple evaluate program that works fine as long as the input is something like '(5 + 4). However I want to be able to solve other inputs such as '(5 +( 3 - 1)) and '(6 + 5) - (4 /2 ) . My problem obviously being how to handle the parentheses. I tried comparing the literal value of '( as in ((equal (car L) '( ) (solve(cdr L))) but that only throws all of my close parentheses

问题 exception div; fun f(x,y) = let val before = 2.0 * x + 3.0 * y in (before + (1.0 / (if x > 0.0001 then x else raise div)) + 2.0 / y) handle div => before / 6.0 end This code yields some compile error. That is e.sml:4.8-4.14 Error: expression or pattern begins with infix identifier "before" e.sml:6.8-6.14 Error: expression or pattern begins with infix identifier "before" e.sml:6.57-6.60 Error: expression or pattern begins with infix identifier "div" e.sml:6.81-6.84 Error: expression or pattern

问题 After some google search I find it! Prefix to Infix This algorithm is a non-tail recursive method. The reversed input string is completely pushed into a stack. prefixToInfix(stack) 1) IF stack is not empty a. Temp -->pop the stack b. IF temp is a operator i. Write a opening parenthesis to output ii. prefixToInfix(stack) iii. Write temp to output iv. prefixToInfix(stack) v. Write a closing parenthesis to output c. ELSE IF temp is a space -->prefixToInfix(stack) d. ELSE i. Write temp to output

问题 My lecturer gave me an assignment to create a program to convert and infix expression to postfix using Stacks. I've made the stack classes and some functions to read the infix expression. But this one function, called convertToPostfix(char * const inFix, char * const postFix) which is responsible to convert the inFix expression in the array inFix to the post fix expression in the array postFix using stacks, is not doing what it suppose to do. Can you guys help me out and tell me what I'm