operators

问题 This is a best practice question. I've only seen the Perl spaceship operator (<=>) used in numeric sort routines. But it seems useful in other situations. I just can't think of a practical use. Can anyone give me an example of when it could be used outside of a Perl sort? 回答1: I'm writing a control system for robot Joe that wants to go to robot Mary and recharge her. They move along the integer points on the line. Joe starts at $j and can walk 1 meter in any direction per time unit. Mary

问题 This is a best practice question. I've only seen the Perl spaceship operator (<=>) used in numeric sort routines. But it seems useful in other situations. I just can't think of a practical use. Can anyone give me an example of when it could be used outside of a Perl sort? 回答1: I'm writing a control system for robot Joe that wants to go to robot Mary and recharge her. They move along the integer points on the line. Joe starts at $j and can walk 1 meter in any direction per time unit. Mary

问题 This is a best practice question. I've only seen the Perl spaceship operator (<=>) used in numeric sort routines. But it seems useful in other situations. I just can't think of a practical use. Can anyone give me an example of when it could be used outside of a Perl sort? 回答1: I'm writing a control system for robot Joe that wants to go to robot Mary and recharge her. They move along the integer points on the line. Joe starts at $j and can walk 1 meter in any direction per time unit. Mary

问题 I am just making sure I understand this concept correctly. With the * operator, I make a new variable, which is allocated a place in memory. So as to not unnecessarily duplicate variables and their values, the & operator is used in passing values to methods and such and it actually points to the original instance of the variable, as opposed to making new copies...Is that right? It is obviously a shallow understanding, but I just want to make sure I am not getting them mixed up. Thanks! 回答1:

问题 Probably a little bit broad question, but the official documentation doesn't even mentioning the arrow operator (or language construct, I don't know which phrase is more accurate) as an independent entity. The most obvious use is the when conditional statement, where it is used to assign an expression to a specific condition: val greet = when(args[0]) { "Appul" -> "howdy!" "Orang" -> "wazzup?" "Banan" -> "bonjur!" else -> "hi!" } println(args[0] +" greets you: \""+ greet +"\"") What are the

问题 Probably a little bit broad question, but the official documentation doesn't even mentioning the arrow operator (or language construct, I don't know which phrase is more accurate) as an independent entity. The most obvious use is the when conditional statement, where it is used to assign an expression to a specific condition: val greet = when(args[0]) { "Appul" -> "howdy!" "Orang" -> "wazzup?" "Banan" -> "bonjur!" else -> "hi!" } println(args[0] +" greets you: \""+ greet +"\"") What are the

问题 I was taking this JavaScript Quiz and found this question - "1" - - "1"; The result of this statement is 2 . Can anyone explain what is going on here? I also found that with even - the addition of strings take place but with odd - subtraction. This only happens when a number is a string. Here are some more eamples- "1" - "1" => 0 "1" - - "1" => 2 "1" - - - "1" => 0 "1" - - - - "1" => 2 "a" - "b" => NaN 回答1: The expression is equivalent to "1" - (-"1") . The unary minus will convert its

问题 I have trouble understanding why python returns different results for these 2 statements: -1 // 3/4 and -1 // 0.75 The first one returns -0.25 and the second on returns -2 . The way i understand it, the / operator is executed before // , thus those 2 statements should have the same result. edit: I was referring to a document provided by my university. I misinterpreted that. Official python documentation proves me wrong. Thanks for all the quick answers. 回答1: The / and // operators have the

问题 So I've been trying to write a calculator with Scala's parser, and it's been fun, except that I found that operator associativity is backwards, and that when I try to make my grammar left-recursive, even though it's completely unambiguous, I get a stack overflow. To clarify, if I have a rule like: def subtract: Parser[Int] = num ~ "-" ~ add { x => x._1._1 - x._2 } then evaluating 7 - 4 - 3 comes out to be 6 instead of 0. The way I have actually implemented this is that I am composing a binary

问题 We can define intrinsic operators of Python as stated here. Just for curiosity, can we define new operators like $ or *** ? (If so, then we can define ternary condition operators or rotate operators.) 回答1: Expanding on @fasouto answer, but adding a bit more code. While you cannot define new operators AND you cannot redefine existing operators for built-in types, what you can do is to define a class (instantiated to any valid Python name, e.g. op ) that act as an intermediate binding for two