operator-precedence

问题 At what point does JavaScript determine the left-hand side of an assignment — is it before or after the right-hand side is evaluated? For example, what does this code do? var arr = [{thing:1},{thing:2},{thing:3},{last:true}]; arr[arr.length - 1].newField = arr.pop(); 回答1: The left-hand side of an assignment operator is evaluated first. The specification for this as of ES2015 can be found in the "Runtime Semantics: Evaluation" portion of the "Assignment Operators" section and can be very

问题 Suppose I have two custom infix operators in R: %foo% and %bar% . I have expressions that use both operators, such as: x %foo% y %bar% z How can I determine the operator precedence of %foo% and %bar% ? How can I change the precedence so that, for example, %bar% always executes before %foo% ? In the example above this would be the same as: x %foo% (y %bar% z) 回答1: I don't think this is explicitly documented, but implicit in the R language documentation is that infix operators are all of equal

问题 If I have the following ... a OrElse b ... and a is True then clearly b is never evaluated. But if I add an Or , then what? a OrElse b Or c Does/should c get evaluated? And what if I put in some brackets? Apologies if this is basic. Of course I can test for the answer myself but I can't find this question answered here or elsewhere. Lots of questions dealing with Or versus OrElse but nothing dealing with Or with OrElse 回答1: This is an operator precedence problem. The relevant documentation is

问题 Let's say we have class A : class A { public: A& func1( int ) { return *this; } A& func2( int ) { return *this; } }; and 2 standlone functions: int func3(); int func4(); now in this code: A a; a.func1( func3() ).func2( func4() ); is order of evaluation of functions func3() and func4() defined? According to this answer Undefined behavior and sequence points one of the sequence points are: at a function call (whether or not the function is inline), after the evaluation of all function arguments

问题 What is the correct sequence of the math operations in this expression in Java: a + b * c / ( d - e ) 1. 4 1 3 2 2. 4 2 3 1 I understand that result is the same in both answers. But I would like to fully understand the java compiler logic. What is executed first in this example - multiplication or the expression in parentheses? A link to the documentation that covers that would be helpful. UPDATE: Thank you guys for the answers. Most of you write that the expression in parentheses is

问题 This question already has answers here : Closed 8 years ago . Possible Duplicates: Output of multiple post and pre increments in one statement Post-increment and pre-increment in 'for' loop The following code snippet int i=0; printf("%d %d",i++,i++); gives the output 1 0 I can understand that, but the following int i=0; printf("%d %d",++i,++i); gives the output 2 2 Can someone explain me the second behavior? 回答1: Both printfs invoke undefined-behavior. See this : Undefined behavior and

问题 In this XPath expression: //div[@id=”myID”]|p , does the // operator get applied to both sides of the union operator? Or would this expression simply return all div elements in the document that have an id attribute value of myID and all p elements that are children of the context node? Is there a reference for XPath operator binding and associativity? 回答1: XPath Operator Order Precedence The XPath EBNF grammar implies the following precedence among operators ( lowest to highest ): Source :

问题 Will right side always evaluated to ahead of left side? And then the result of right side will be passed on to left side. I am not talking about the exception such as A[i]=i++ I am talking about the normal cases: A[i] = (j+32+43 & K); A[j] != (A[j] + A[k]); will the right part of all these expression evaluated first and then the result is compared to the left side? (Always) 回答1: In general the order of evaluation of sub-expressions is unspecified, there are a few exceptions such as logical

问题 I was just bitten by the following scenario: >>> -1 ** 2 -1 Now, digging through the Python docs, it's clear that this is intended behavior, but why? I don't work with any other languages with power as a builtin operator, but not having unary negation bind as tightly as possible seems dangerously counter-intuitive to me. Is there a reason it was done this way? Do other languages with power operators behave similarly? 回答1: That behaviour is the same as in math formulas, so I am not sure what

问题 This question already has answers here : Is the right-hand side of an assignment always evaluated before the assignment? (2 answers) Closed 4 years ago . In other words is d = {} d["key"] = len(d) safe in Python? I know this is undefined behaviour in C++; where the program might get a reference to the element before computing the value it's going to assign to it. Is this similar in Python or is len(d) always computed before d.__getitem__("key") ? 回答1: Yes, in Python it is safe: the evaluation