parentheses

问题 I've been working on a bit of code to take a bunch of histograms from other files and plot them together. In order to make sure the legend displays correctly I've been trying to take the titles of these original histograms and cut out a bit of information that isn't needed any more. The section I don't need takes the form (A mass=200 GeV), I've had no problem removing what's inside the parentheses, unfortunately everything I've tried for the parentheses themselves either has no effect,

问题 For example if the parenthesis/brackets is matching in the following: ({}) (()){}() () and so on but if the parenthesis/brackets is not matching it should return false, eg: {} ({}( ){}) (() and so on. Can you please check this code? Thanks in advance. public static boolean isParenthesisMatch(String str) { Stack<Character> stack = new Stack<Character>(); char c; for(int i=0; i < str.length(); i++) { c = str.charAt(i); if(c == '{') return false; if(c == '(') stack.push(c); if(c == '{') { stack

问题 I am building a expression analyser from which I would like to generate database query code, I've gotten quite far but am stuck parsing BinaryExpressions accurately. It's quite easy to break them up into Left and Right but I need to detect parenthesis and generate my code accordingly and I cannot see how to do this. An example [please ignore the flawed logic :)]: a => a.Line2 != "1" && (a.Line2 == "a" || a.Line2 != "b") && !a.Line1.EndsWith("a") I need to detect the 'set' in the middle and

问题 This question already has answers here : What exactly is a type cast in C/C++? (4 answers) Closed 4 years ago . decimal d = 2; int i = (int) d; I've seen this several times in which parentheses are wrapped around data types. Why not just use int i = int d; ? 回答1: The usage of (int) is called cast ing (or, type-casting). It is essentially telling that, interpret convert the value of d as to an int (integer) and store it into i . In other words, it is a way of converting a type to another one

问题 I have to split a string and I want to avoid splitting it with commas inside parentheses. So how can I implement that? Example: $string = "string1 (sString1, sString2,(ssString1, ssString2)), string2, string3"; result should be: array( [0] => string1 (sString1, sString2,(ssString1, ssString2)) [1] => string2 [2] => string3 ) 回答1: You're going to get in an awful mess trying to do this with regex. It's very simple to loop through the characters of a string and do this kind of checking.

问题 First, define a string of "balanced" parentheses as a string such that for every '(' there is one unique, matching ')' somewhere after that '('. For example, the following strings are all "balanced": () ()() (())() while these are not: )( ())( Given a string of parentheses (length <= 1,000,000) and a list of range queries, find the maximum length of balanced parentheses within each of the ranges for each of the <= 100,000 queries (using 0-indexing for the ranges) Ex: ()))()(()) Range: [0,3] -

问题 Sorry, but this is a newbie question. There are also a lot of questions on "link_to" already, but none answer my question, which I don't think is really specific to link_to ... Creating a link like this works : <%= link_to person.automobile_id, person %> But, trying it like this does not work : <%= link_to (translation.request_id, translation) %> This produces an error: syntax error, unexpected ',', expecting ')' This confuses me, since to me it looks like I'm just wrapping parenthesis around

问题 I just spent 3 hours wondering why I couldn't start a session, then realised that I used: session_start; when I should have used: session_start(); I received no error messages at all! Perhaps then, I thought, it's a lazy way to differentiate functions from variables - but then remembered that it can't be as variables require a $ Can anyone tell me why parentheses are required then, and also why no error is given when they aren't used? 回答1: I get the notice (Having set error_reporting to E_ALL

问题 (first time poster, long time visitor via Google) I'm trying to extract the contents of some square brackets, however i'm having a spot of bother. I've got it working for round brackets as seen below, but I can't see how it should be modified to work for square brackets. I would have thought replacing the round for square and vice versa in this example should work, but apparently not. It needs to ignore brackets within brackets. So it won't return (11) but will return (10(11)12). $preg = '#\(

问题 As you see in the screen shot above, I have the following expression in my Matlab m-file code: K = P * D * D' * P; Where, P is an nxn matrix, and D is a nx1 column vector (n=4, if it matters). Why am I getting this warning message? What changes if I use or don't use parenthesis there? 回答1: Floating-point arithmetic is not associative. So in general, a * (b * c) won't necessarily give the same result as (a * b) * c . Your statement as written is equivalent to ((P * D) * D') * P , so the