Question

为什么调用fn（Iterator（“foo”）编译，但对fn（fooIterator）的调用失败并显示错误“ type mismatch; found：Iterator [java.lang.String] required：scala .Iterator [com.banshee.Qx.HasLength] “

object Qx {
    type HasLength = {def length: Int}
    def fn(xs: Iterator[HasLength]) = 3
    var tn = fn(Iterator("foo"))
    var fooIterator = Iterator("foo")
    var tnFails = fn(fooIterator) //doesn't compile
}

他们不是一回事吗？

Answer 1

这个表述有效：

object Qx {
    type HasLength = {def length: Int}
    def fn[HL <% HasLength](xs: Iterator[HL]) = 3
    val tn = fn(Iterator("foo"))
    val fooIterator = Iterator("foo")
    val tnFails = fn(fooIterator)
}

Answer 2

它必须是改进表示中的错误，因为以下两个公式都有效。

object Qx1 {
    // give length() parens, even though the Iterator definition doesn't have them
    type HasLength = { def length(): Int }

    def fn(xs: Iterator[HasLength]) = 3
    var tn = fn(Iterator("foo"))
    var fooIterator = Iterator("foo")
    var tnFails = fn(fooIterator) //doesn't compile
}

object Qx2 {
    type HasLength = { def length: Int }

    def fn(xs: Iterator[HasLength]) = 3
    var tn = fn(Iterator("foo"))
    // annotate the type of fooIterator before the type inferencer can mis-infer
    var fooIterator: Iterator[HasLength] = Iterator("foo")
    var tnFails = fn(fooIterator) //doesn't compile
}

编辑：

早上太早了。它是带有length（）方法的String，它有parens，这意味着它是正确的，你认为length和length（）是相同的方法。（这是我以前记录过的一个很好的小陷阱。）

这两个调用函数采用结构类型集合之间的区别是什么？

2 个答案: