operator-precedence

问题 Unthinkingly I wrote some code to check that all the values of a struct were set to 0. To accomplish this I used: bool IsValid() { return !(0 == year == month == day == hour == minute == second); } where all struct members were of type unsigned short. I used the code as part of a larger test but noticed that it was returning false for values differing from zero, and true for values that were all equal to zero - the opposite of what I expected. I changed the code to read: bool IsValid() {

问题 Given the following example: declare @i int select @i = 1, @i = 2 select @i Will @i always be 2? This is about the most trivial example I can think of, but I am considering using this for swapping values in variables. I also believe this method of assignment (select) is not ANSI compliant (however useful), but don't really care to have portable code in this case. UPDATE Thanks to @MichaelFredrickson, we have @MartinSmith's answer and reference to MSDN on this. I am now struggling with what

问题 Last time I got confused by the way PowerShell eagerly unrolls collections, Keith summarized its heuristic like so: Putting the results (an array) within a grouping expression (or subexpression e.g. $()) makes it eligible again for unrolling. I've taken that advice to heart, but still find myself unable to explain a few esoterica. In particular, the Format operator doesn't seem to play by the rules. $lhs = "{0} {1}" filter Identity { $_ } filter Square { ($_, $_) } filter Wrap { (,$_) }

问题 The following code snippet: int i=-3,j=2,k=0,m; m=++i && ++j || ++k; can be evaluated using two concepts,I believe: 1.Since ++ operator has greater precedence than the logical operators,so first all increment operators will be evaluted,then && having higher precedence than || will be computed.In this process,k will be incremented. 2.First && operator will be evaluated.For this ++ i and ++j will be computed.Since the result of the && operator is 1,no need to evaluate the ++k.So k will not be

问题 For example, in JavaScript we could write a program like this: var a = 1; testFunction(++a, ++a, a); function testFunction(x, y, z){ document.writeln("<br />x = " + x); document.writeln("<br />y = " + y); document.writeln("<br />z = " + z); } and we would get an output: x = 2 y = 3 z = 3 This implies that parameters are truly evaluated from left to right in JavaScript. In C we would get output x = 3 y = 3 z = 3 I was wondering if we could do the same in Python or is it impossible since it's a

问题 In Javascript certain operators are processed before others: 1 + 2 * 3 // 1 + (2 * 3) // 7 because * has higher precedence than + 1 === 0 + 1 // 1 === (0 + 1) // true because + has a higher precedence than === The MDN has a full breakdown of all operators and their precedence ... except await . await getFoo() * 2; // await (getFoo() * 2) or (await getFoo()) * 2? await getFoo() === 5; // await (getFoo() === 5) or (await getFoo()) === 5? Can anyone explain which operators are processed before

问题 I have a test in Excel VBA: If (test1) And (test2) And (test3) Then 'do something End If In C, Java, etc. test1 would be run first, then test2, then test3. Critically, if test1 is false the whole test is false so the remaining tests do not run. Does that happen in this case with VBA? If so, in which order are the tests running? 回答1: In all VBs prior to .NET there is no such thing as short-circuit. All expressions will be evaluated even if not required. If you want short-curcuit, do nested IFs

问题 int it=9, at=9; if(it>4 || ++at>10 && it>0) { System.out.print("stuff"); } System.out.print(at); prints out stuff9 and I want to know why as I thought ++at>10 && it>0 would be evaluated first and thus make at = 10. 回答1: Operator precedence only controls argument grouping; it has no effect on execution order. In almost all cases, the rules of Java say that statements are executed from left to right. The precedence of || and && causes the if control expression to be evaluated as it>4 || (++at

问题 What is the standard behaviour of a session variable when used in an SQL statement. Case 1 : In the following example, session variable is behaving as expected. mysql> set @m1=0, @m2=0, @m3=0; Query OK, 0 rows affected (0.00 sec) mysql> mysql> select -> @m1 := 55 m1, @m2 := 42 m2, @m3 := 66 m3, -> @m1, @m2, @m3, -> @b1 := greatest( @m1, @m2, @m3 ) b1, -> @b2 := ( ( @total := @m1 + @m2 + @m3 ) -> - ( @b1 + least( @m1, @m2, @m3 ) )) b2, -> @total total; +----+----+----+------+------+------+----

问题 Why does the first line return TRUE, and the third line returns 1? I would expect both lines to return 1. What is the exact meaning of those extra two parentheses in the third line? !is.na(5) + !is.na(NA) # TRUE (!is.na(5)) + (!is.na(NA)) # 1 edit: should check these multiple times. The original problem was with !is.na() , thought it replicated for is.na() . But it didn't :) 回答1: ! has a weird, counter-intuitive precedence in R. Your first code is equivalent to !(is.na(5) + !is.na(NA)) That