operator-precedence

问题 To my knowledge, C++ always evaluates from left to right in a conditional statement if(A, B, C) A would be evaluated first, B second, so on. However, the following example is exhibiting some odd behavior. #include <iostream> #include <map> #include <memory> #include <vector> #include <boost/logic/tribool.hpp> //-//////////////////////////////////////////// // File Block class FileBlock { public: FileBlock(); virtual ~FileBlock(); bool linked(); std::vector<int> messages_; private: boost:

问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 6 years ago . Why not simply evaluate Left to Right? Can someone explain how precedence makes code more readable? To me it seems to require more

问题 Code: #include<stdio.h> int main() { int j = 7, i = 4; j = j || ++i && printf("you can"); printf("%d %d",i,j); return 0; } Output: 4 1 [Code Link][1] The precedence of prefix operator is higher than logical operators. 2.Logical && has higher precedence than logical || . In Logical AND( && ) if first operand evaluates to false than second will not be evaluated and In Logical OR( ||) if first operand evaluates to true , then second will not be evaluated. The complete expression is evaluating to

问题 I'm trying to write more efficient code in my scripts and have been implementing ternary conditional operators on occasion. I can't understand why I am getting an additional result when using a ternary conditional operator in a loop: #!/usr/bin/perl use strict; use warnings; my @array = ('Serial = "123"', 'Serial = "456"', 'Serial = "789"'); my ($test1,$test2); foreach my $a (@array){ !$test1 ? $test1 = $a : $test1 .= " AND " . $a; } foreach my $b (@array){ if (!$test2) { $test2 = $b } else {

问题 Wikipedia claims that the [] operator precedes the * operator in evaluation. Then, why does the following statement: char *a[3]; declare an array of 3 character pointers, rather than a pointer to an array of 3 characters as per the operator precedence? 回答1: Because, as Wikipedia says, [] has higher precedence than * ? Processing the declaration, the a[3] is processed as 'array of 3' before you process the * . To declare a pointer to an array of three characters, you have to use parentheses to

问题 I'm trying to write a String evaluation function i.e. evaluate("4 + 1") ; // returns 5 evaluate("4 + 1 + 3") ; // returns 8 evaluate("4 + 1 * 3") ; // returns 7 (not 15) The operators are + - / and * My initial though was to use regular expressions to collect operators and digits as these can be matched. And than after finding that info, somehow figure out a way to prioritize /* ove -+ operators. Here is how I started : static String regex = "([\\+\\*-/])+"; static String digitRegex = "(\\d)+

问题 I have seen: CSS 2, precedence of stylesheets imported using link element In which order do CSS stylesheets override? stylesheet - Can one CSS file take priority over another CSS file? They all seem to say that, for the same selector occurring multiple times, the last one wins. However that is not what happens for me. So given that "Aqua.css" has: body {color:aqua;} And "Red.css" has: body {color:red;} Then using the following: <link rel="stylesheet" type="text/css" href="Aqua.css" title=

问题 For the following code snippet I get the output as 1 . I want to know how it came? void main() { int x=10,y=20,z=5,i; i=x<y<z; printf("%d",i); } 回答1: i=x<y<z; , gets interpreted as i=(x<y)<z , which in turn gets interpreted as i=1<z , which evaluates to 1. 回答2: 10 is less than 20, resulting in 1, and 1 is less than 5, resulting in 1. C doesn't chain relational operators as some other languages do. 回答3: It operates as follows: Since < is a logical expression, x<y i.e 10<20 is true i.e 1. So it

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Problem with operator precedence we know that precedence of prefix is greater than "LOGICAL AND" ( && ) and precedence of "LOGICAL AND" is greater than "LOGICAL OR" ( || ). Below program seems to violate it: int main() { int i=-3,j=2,k=0,m; m=++i||++j&&++k; printf("%d %d %d %d",i,j,k,m); return 0; } If precedence of ++ is more than && and || then all prefix should execute first. After this i=-2,j=3,k=1 and then

问题 As part of learning Ruby am trying to implement a basic interpreter which reads input and do basic arithmetic calculations. So far basic arithmetic operations are working but having problem in operator precedence. Which is not handled yet. This is the code. Am at a beginner level. Any mistakes in this code are due to my lack of knowledge. How this code can be modified to handle operator precedence. Sample output 2+2+2 = 6 #correct 10+10/2 = 10 # incorrect as in irb answer must be 15 Github