带有两个重载方法编译的示例代码(Scala 2.11.8),如预期的那样:
def foo(p: Int, t: Double): Unit = {}
def foo(r: Int => Double): Unit = {}
foo(0, 0) // 1st method
foo(x => x.toDouble) // 2nd method
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添加了通用参数的相同代码:
def fooGeneric[T](p: T, t: Double): Unit = {}
def fooGeneric[T](r: T => Double): Unit = {}
fooGeneric[Int](1, 1) // 1st method
fooGeneric[Int]((x: Int) => x.toDouble) //2nd method
fooGeneric[Int](x => x.toDouble) // does not compile
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产生以下编译错误:
Error:(112, 19) missing parameter type
fooGeneric[Int](x => x.toDouble)
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方法解析似乎没有歧义,但编译器无法找到匹配项.为什么?
def test[T: ClassTag]: T = {
println(classTag[T])
null.asInstanceOf[T]
}
val x1: Int = test
val x2: Int = test[Int]
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印刷
Nothing
Int
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我希望编译器能够猜测Int类型,而无需显式提供它(编辑:在右侧,即使工作val x1: Int = test)。
是否有解决方法可以摆脱显式类型注释?