operator-precedence

问题 I have three classes all of which needs to be enabled/disabled based on the @ConditionalOnExpression annotaion. All of these are in separate .java files. EX: @ConditionalOnExpression("#T(com.xxx.xxx.xxx.xxx.CxProperties).isAEnabled()}") Class A{ } @ConditionalOnExpression("#T(com.xxx.xxx.xxx.xxx.CxProperties).isBEnabled()}") Class B{ } @ConditionalOnExpression("#T(com.xxx.xxx.xxx.xxx.CxProperties).isCEnabled()}") Class C{ } Now i have an init funtion in a different class which gets executed

问题 This question already has answers here : Why does x++ have higher precedence than ++x? (2 answers) Why does postfix operator++ have higher precedence than prefix operator++? (3 answers) Closed 5 years ago . I understand how post-incrementing and pre-incrementing in Java works, and I've read countless other threads/questions from other people regarding this same topic. BUT! I have yet to learn WHY post-incrementing has higher precedence than pre-incrementing in Java. Does anyone know? To my

问题 I have an application which detects pressed keys on C# winform. The problem is on client side there are some other applications which also detects the pressed keys and suppresses them and because of that my App's form does not get the notification for the pressed key. I have used traditional form.KeyDown event to detect the key. How can I take higher precedence than the other application? 回答1: If you're worried about other apps basically stealing your keystroke then you need to look into

问题 I try to implement a function call in a custom grammar (plus a similar array-access operator): expression : ....OTHER EXPRESSION RULES.... | expression PARENTHESIS_OPEN expressions PARENTHESIS_CLOSE { } %prec DOT | expression SQUARE_OPEN expressions SQUARE_CLOSE { } %prec DOT ; Here is my all my operator precedences: %right ASSIGN ASSIGN_MOD ASSIGN_XOR ASSIGN_AND ASSIGN_STAR ASSIGN_MINUS ASSIGN_PLUS ASSIGN_OR ASSIGN_DIV ASSIGN_LSHIFT ASSIGN_RSHIFT %right QUESTION COLON %left OR %left AND

问题 This question already has answers here : Is short-circuiting logical operators mandated? And evaluation order? (7 answers) Closed 5 years ago . (this question is an exact copy of Is compound if checking for null and then other condition in C++ always safe? but about C , not C++. It was pointed out that the question should be more specific). I have been using the following type of if condition for a lot of time. char* ptr = ...; if (ptr != NULL && ptr[0] != '\0') // <=== is this always safe? {

问题 This question already has answers here : 'AND' vs '&&' as operator (10 answers) Closed 4 years ago . Why is $x true in this statment? $x = true and false; I've got the problem with some variables but I could reduce the problem to the primitive boolean values. Update: As you see in the replies the effect has to do with the operator precedense in PHP. There is also a good explanation about the problem in this question, which I couldn't find in the net before since I didn't know that I have a

问题 This question already has answers here : Is short-circuiting logical operators mandated? And evaluation order? (7 answers) Closed 5 years ago . (this question is an exact copy of Is compound if checking for null and then other condition in C always safe? but about C++ , not C. It was pointed out that the question should be more specific). I have been using the following type of if condition for a lot of time. char* ptr = ...; if (ptr != NULL && ptr[0] != '\0') // <=== is this always safe? { /

问题 If only the second operand in addition is a string then the output is a string: let a = 1 + '2'; console.log(typeof a); // string And if, instead, only the first operand is a string, then the output is still a string: let b = '1' + 2; console.log(typeof b); // string I was guessing that there would be some kind of argument precedence. Is there a reason why this mathematical function defaults to a non-numerical output with mixed-type arguments? 回答1: As it is often the case, the answer is

问题 The C++11 standard (5.17, expr.ass) states that In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression. With respect to an indeterminately-sequenced function call, the operation of a compound assignment is a single evaluation Does this mean, that the expression: int a = 1, b = 10; int c = (a+=1) + (b+=1); if ( c == 10+1+1+1 ) { printf("this is guaranteed"); } else { printf("not

问题 could somebody explain the difference between these two codes? bad_command 2>& >> file.out and bad_command >> file.out 2>& The manual said that these two codes are different,and first command will output nothing to file.out. So , here are my questions. 1/ what is the reason for that? 2/ Is there is a document which describes how operator precedence works in shell? how shell parses and made it's syntax tree. 3/ What is the correct syntax and order of it? --Thanks in advance-- 回答1: Both are