Scala的行为/理解数值类型的隐式变换？

Question

我试图理解Scala for-loop隐式盒子/拆箱"数字"类型的行为.为什么这两个首先失败但其余的失败？

1)失败:

scala> for (i:Long <- 0 to 10000000L) {}


      <console>:19: error: type mismatch;<br>
      found   : Long(10000000L)
      required: Int
              for (i:Long <- 0 to 10000000L) {}
                                  ^

scala> for (i <- 0 to 10000000L) {}

      <console>:19: error: type mismatch;
      found   : Long(10000000L)
      required: Int
          for (i <- 0 to 10000000L) {}
                         ^

scala> for (i:Long <- 0L to 10000000L) {}

scala> for (i <- 0L to 10000000L) {}

scala> for (i:Long <- 0 to 10000000L) {}


      <console>:19: error: type mismatch;<br>
      found   : Long(10000000L)
      required: Int
              for (i:Long <- 0 to 10000000L) {}
                                  ^

4)作品:

scala> for (i <- 0 to 10000000L) {}

      <console>:19: error: type mismatch;
      found   : Long(10000000L)
      required: Int
          for (i <- 0 to 10000000L) {}
                         ^

Answer 1

它与for循环无关:

0 to 1L   //error
0 to 1    //fine
0L to 1L  //fine
0L to 1   //fine

这只是因为to可用的方法Int需要一个Int作为其参数.因此,当你给它一个Long它不喜欢它,你会得到一个错误.

以下是该to方法的定义,可在RichInt以下位置找到:

def to(end: Int): Range.Inclusive = Range.inclusive(self, end)