在以下代码中,隐式转换应用于println(2)行;我愚蠢地期望它适用于整个区块{println(1); println(2)}.我应该如何推断编译器放置隐含的位置？
object Executor {
private var runnable: Runnable = _
def setRunnable(runnable: Runnable) {
this.runnable = runnable
}
def execute() { runnable.run() }
}
object Run extends App {
implicit def blockToRunnable(p: ⇒ Any): Runnable =
new Runnable { def run() = p }
Executor.setRunnable {
println(1)
println(2)
}
println("Before execute")
Executor.execute()
}
解决方法
根据规范,当表达式的类型与期望的类型不匹配时,将应用隐式转换.关键的观察是在键入块时如何线程化预期的类型.
if an expression
eis of typeT,and T does not conform to the expression’s expected typept. In this case an implicit v is searched which is applicable to e and whose result type conforms to pt.
在第6.11节块中,块的最后一个表达式的预期类型定义为
The expected type of the final expression
eis the expected type of the block.
鉴于此规范,编译器似乎必须采用这种方式.块的预期类型是Runnable,println(2)的预期类型也变为Runnable.
建议：如果您想知道应用了哪些implicits,可以使用适用于Eclipse的Scala IDE 2.1的每晚构建.它可以“突出隐含”.
编辑：我承认,当范围内隐含着一个名字叫做时,这是令人惊讶的.
