Question

为什么要编造假(Iterator(“foo”),但打耳机(fooIterator)的电话却有误,“

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
}

更不用说什么?

Answer 1

这必须成为改进的体现,因为以下两条表述都是工作。

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
}

Edit:

上午太早。这套方法有长处,这意味着它正确,而且你在思考时间长度和时间长度方面有错误。 (这只剩下一个小小小小块的陷阱,在前有记载。)

Answer 2

这一表述有助于:

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)
}

友情链接