operator-precedence

问题 This question already has answers here : Evaluation of the following expression (3 answers) Closed 3 months ago . As I know logical operator && has higher precedence than || . On running the code: #include <stdio.h> int main() { int i = 1, j =1, k = 1; printf(\"%d\\n\",++i || ++j && ++k); printf(\"%d %d %d\",i,j,k); return 0; } is giving the output: 1 2 1 1 which is possible only when ++i || ++j && ++k is evaluated like this: (++i) || (++j && ++k) But, according to operator precedence rule it

问题 In the expression a + b , is a guaranteed to be evaluated before b , or is the order of evaluation unspecified? I think it is the latter, but I struggle to find a definite answer in the standard. Since I don\'t know whether C handles this different from C++, or if evaluation order rules were simplified in C++11, I\'m gonna tag the question as all three. 回答1: In C++, for user-defined types a + b is a function call, and the standard says: §5.2.2.8 - [...] The order of evaluation of function

问题 The Python documentation for operator precedence states: Operators in the same box group left to right (except for comparisons, including tests, which all have the same precedence and chain from left to right — see section Comparisons...) What does this mean? Specifically: \"Operators in the same box group left to right (except for comparisons...)\" -- do comparisons not group left to right? If comparisons do not group left to right, what do they do instead? Do they \"chain\" as opposed to \

问题 I dont understand how that output (\" four \") comes? $a = 2; echo $a == 1 ? \'one\' : $a == 2 ? \'two\' : $a == 3 ? \'three\' : $a == 5 ? \'four\' : \'other\' ; // prints \'four\' I don\'t understand why \" four \" gets printed. 回答1: You need to bracket the ternary conditionals: <?php for ($a=0; $a < 7; $a++) { echo ( $a == 1 ? 'one' : ($a == 2 ? 'two' : ($a == 3 ? 'three' : ($a == 5 ? 'four' : 'other')))); echo "\n"; // prints 'four' } exit; ?> returns: other one two three other four other

问题 Who likes to tell me what is wrong with this code (syntactically)? -- merge two sorted lists mergeX [] b res = b ++ res mergeX a [] res = a ++ res mergeX a:as b:bs res | a > b = mergeX as b:bs a:res | otherwise = mergeX a:as bs b:res Interpreter: Parse error in pattern: mergeX 回答1: You need some parenthesis: mergeX [] b res = b ++ res mergeX a [] res = a ++ res mergeX (a:as) (b:bs) res | a > b = mergeX as (b:bs) (a:res) | otherwise = mergeX (a:as) bs (b:res) The reason is because : has a

问题 Misunderstanding Java operator precedence is a source of frequently asked questions and subtle errors. I was intrigued to learn that even the Java Language Specification says, \"It is recommended that code not rely crucially on this specification.\" JLS §15.7 Preferring clear to clever , are there any useful guidelines in this area? Here are a number of resources on the topic: JLS Operators JLS Precedence Java Glossary Princeton Oracle Tutorial Conversions and Promotions Java Operator

问题 Suppose I have a number of statements that I want to execute in a fixed order. I want to use g++ with optimization level 2, so some statements could be reordered. What tools does one have to enforce a certain ordering of statements? Consider the following example. using Clock = std::chrono::high_resolution_clock; auto t1 = Clock::now(); // Statement 1 foo(); // Statement 2 auto t2 = Clock::now(); // Statement 3 auto elapsedTime = t2 - t1; In this example it is important that the statements 1

问题 In the C programing language, why do the bitwise operators (& and |) have lower precedence than the equality operator (==)? It does not make sense to me. 回答1: You need to ask Brian Kernighan or Dennis Ritchie. From this forum: http://bytes.com/topic/c/answers/167377-operator-precedence The && and || operators were added later for their "short-circuiting" behavior. Dennis Ritchie admits in retrospect that the precedence of the bitwise operators should have been changed when the logical

问题 I have read this in answer to many questions on here. But what exactly does it mean? var test = new Dictionary<int, string>(); test.Add(0, \"zero\"); test.Add(1, \"one\"); test.Add(2, \"two\"); test.Add(3, \"three\"); Assert(test.ElementAt(2).Value == \"two\"); The above code seems to work as expected. So in what manner is a dictionary considered unordered? Under what circumstances could the above code fail? 回答1: Well, for one thing it's not clear whether you expect this to be insertion-order

问题 I am wondering how or/and works? For example if I want to get all rows where display = 1 I can just do WHERE tablename.display = 1 and if I want all rows where display = 1 or 2 I can just do WHERE tablename.display = 1 or tablename.display = 2 But what if I want to get all rows where display = 1 or 2 and where any of the content, tags, or title contains hello world How would the logic play out for that? Select * from tablename where display = 1 or display = 2 and content like \"%hello world%\