为什么 Scala 的分号推断在这里失败? [英] Why does Scala's semicolon inference fail here?

问题描述

在使用 Scala 2.7.3 编译以下代码时，

On compiling the following code with Scala 2.7.3,

package spoj

object Prime1 {
  def main(args: Array[String]) {
    def isPrime(n: Int) = (n != 1) && (2 to n/2 forall (n % _ != 0))
    val read = new java.util.Scanner(System.in)
    var nTests = read nextInt // [*]
    while(nTests > 0) {
      val (start, end) = (read nextInt, read nextInt)
      start to end filter(isPrime(_)) foreach println
      println
      nTests -= 1
    }
  }
}

我收到以下编译时错误:

I get the following compile time error :

PRIME1.scala:8: error: illegal start of simple expression
    while(nTests > 0) {
    ^
PRIME1.scala:14: error: block must end in result expression, not in definition
  }
  ^
two errors found

当我在注释为 [*] 的行的末尾添加分号时，程序编译正常.任何人都可以解释为什么 Scala 的分号推断无法在该特定行上工作?

When I add a semicolon at the end of the line commented as [*], the program compiles fine. Can anyone please explain why does Scala's semicolon inference fail to work on that particular line?

推荐答案

是不是因为scala 假设你正在使用语法a foo b(相当于a.foo(b)) 在您对 readInt 的调用中.也就是说，它假设 while 循环是 readInt 的参数(回想一下每个表达式都有一个类型)，因此最后一个语句是一个声明:

Is it because scala is assuming that you are using the syntax a foo b (equivalent to a.foo(b)) in your call to readInt. That is, it assumes that the while loop is the argument to readInt (recall that every expression has a type) and hence the last statement is a declaration:

var ntests = read nextInt x

其中x 是你的 while 块.

wherex is your while block.

我必须说，作为一个偏好点，我现在回到使用通常的 a.foo(b) 语法而不是 a foo b 除非专门使用专为这种用途而设计的 DSL(例如演员的 a !b).总的来说，它使事情变得更加清晰，而且您不会被这种奇怪的东西咬伤！

I must say that, as a point of preference, I've now returned to using the usual a.foo(b) syntax over a foo b unless specifically working with a DSL which was designed with that use in mind (like actors' a ! b). It makes things much clearer in general and you don't get bitten by weird stuff like this!

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