infix-notation

Converting from Infix to Postfix and evaluating Postfix notation

戏子无情 提交于 2021-02-07 11:13:50
问题 I'm writing a program that reads an Infix notation, converts it to Postfix and then evaluate that Postfix. Here's my program: #include<stdio.h> #include <ctype.h> #define SIZE 50 /* Size of Stack */ char s[SIZE]; int top = -1; /* Global declarations */ push(char elem) { /* Function for PUSH operation */ s[++top] = elem; } char pop() { /* Function for POP operation */ return (s[top--]); } int pr(char elem) { /* Function for precedence */ switch (elem) { case '#': return 0; case '(': return 1;

Converting from Infix to Postfix and evaluating Postfix notation

只谈情不闲聊 提交于 2021-02-07 11:13:45
问题 I'm writing a program that reads an Infix notation, converts it to Postfix and then evaluate that Postfix. Here's my program: #include<stdio.h> #include <ctype.h> #define SIZE 50 /* Size of Stack */ char s[SIZE]; int top = -1; /* Global declarations */ push(char elem) { /* Function for PUSH operation */ s[++top] = elem; } char pop() { /* Function for POP operation */ return (s[top--]); } int pr(char elem) { /* Function for precedence */ switch (elem) { case '#': return 0; case '(': return 1;

What is easiest way to calculate an infix expression using C language?

落爺英雄遲暮 提交于 2021-02-04 09:29:07
问题 Suppose the user inputs an infix expression as a string? What could be the easiest ( By easiest I mean the shortes t) way to evaluate the result of that expression using C language? Probable ways are converting it to a postfix then by using stacks.But its rather a long process. Is there any way of using functions such as atoi() or eval() that could make the job easier? 回答1: Certainly the most instructive way (and possibly even the easiest, once you know how) is to learn how to write your own

Insert space after every one or two characters in string

时光毁灭记忆、已成空白 提交于 2021-01-27 17:24:49
问题 I need to write code to correct user input. The input is a mathematical expression in infix form. In order for my calculations to be correct, i need each operand spaced away from the operator. Can you please help me write a method that will provide the following expected results. Expected input: 13*21-5+33 Expected Output: 13 * 21 - 5 + 33 Research I have done: I could only find out how to add spaces between every character in the string and that will not work for calculating expressions with

Insert space after every one or two characters in string

余生长醉 提交于 2021-01-27 17:15:12
问题 I need to write code to correct user input. The input is a mathematical expression in infix form. In order for my calculations to be correct, i need each operand spaced away from the operator. Can you please help me write a method that will provide the following expected results. Expected input: 13*21-5+33 Expected Output: 13 * 21 - 5 + 33 Research I have done: I could only find out how to add spaces between every character in the string and that will not work for calculating expressions with

Recursive descent: infix to postfix conversion repeating operators

ε祈祈猫儿з 提交于 2020-06-13 11:45:49
问题 We recently learned about converting infix to postfix using stacks during our programming course at uni. And I have meaning to write my parser for a while, so I decided to it using recursive descent. I am following along with this: Parsing Expressions by Recursive Descent Theodore Norvell . Here is the grammar he uses: E --> P {B P} P --> v | "(" E ")" | U P B --> "+" | "-" | "*" | "/" | "^" U --> "-" | "+" I have attempted at implementing this in C and it works. However If I give it the

Scala Map, ambiguity between tuple and function argument list

风流意气都作罢 提交于 2020-05-14 17:56:50
问题 val m = scala.collection.mutable.Map[String, Int]() // this doesn't work m += ("foo", 2) // this does work m += (("foo", 2)) // this works too val barpair = ("bar", 3) m += barpair So what's the deal with m += ("foo" , 2) not working? Scala gives the type error: error: type mismatch; found : java.lang.String("foo") required: (String, Int) m += ("foo", 2) ^ Apparently Scala thinks that I am trying to call += with two arguments, instead of one tuple argument. Why? Isn't it unambiguous, since I

Infix to postfix conversion in java

匆匆过客 提交于 2020-01-13 18:15:46
问题 I have a java class that converts infix expression to postfix. I have managed to make the class run without errors but it is not giving me the right output. The output is suppose to be in form : InToPost: converting expressions from infix to postfix... [A, +, B] [A, +, B, *, C] [A, +, B, *, C, +, D] [(, A, +, B, ), *, C] [(, A, +, B, ), /, (, C, -, D, )] [(, (, A, +, B, ), *, C, -, (, D, -, E, ), )] InToPost: emitting postfix expressions... A B + A B C * + A B C * + D + A B + C * A B + C D -

Infix to postfix conversion in java

生来就可爱ヽ(ⅴ<●) 提交于 2020-01-13 18:15:29
问题 I have a java class that converts infix expression to postfix. I have managed to make the class run without errors but it is not giving me the right output. The output is suppose to be in form : InToPost: converting expressions from infix to postfix... [A, +, B] [A, +, B, *, C] [A, +, B, *, C, +, D] [(, A, +, B, ), *, C] [(, A, +, B, ), /, (, C, -, D, )] [(, (, A, +, B, ), *, C, -, (, D, -, E, ), )] InToPost: emitting postfix expressions... A B + A B C * + A B C * + D + A B + C * A B + C D -

Converting Infix to prefix notation in JavaScript

别说谁变了你拦得住时间么 提交于 2020-01-06 05:26:18
问题 (I did ask a similar question in the past, but the documentation was wrong, so this is the correct version of that past question) Please help me in JavaScript: The program that I am coding is one that takes in an expression in prefix notation and outputs the same expression in infix notation. The idea behind this program is as follows: if the user enters + 1 2 the expected output is 1 + 2. All valid symbols are +, -, *, /, and %. The amount of numbers that the user can enter should be