scala unapply的测试

object Name {def unapply(ipt:String)={val pos = ipt.indexOf(" ")
    if (pos == -1) None 
    else Some((ipt.substring(0,pos),ipt.substring(pos+1)))
  }
}
object IsCompound{def unapply(ipt:String)=ipt.contains(" ")}
val author = "aabc 111 1"
println(IsCompound.unapply(author))
author match{case Name(f,uu IsCompound())=>println("haha")}
author match{case Name(f,uu IsCompound(t))=>println("haha")}
author match{case Name(f,IsCompound(t))=>println("haha")}
author match{case Name(f,uu@IsCompound(t))=>println("haha")}
author match{case Name(f,uu@IsCompound())=>println("haha")}
/* if:uu IsCompound() : unapply receive uu; so it's a wrong syntax in the case of IsCompound has a boolean unapply 
 * if: uu IsCompound(t) : equals IsCompound(uu,t);so it's a wrong syntax in the case of IsCompound has a boolean unapply 
 * if: IsCompound(t) : equals uu IsCompound();so this it's a wrong syntax in the case of IsCompound has a boolean unapply 
 * if: uu@IsCompound(t) : unapply receive uu and t is set the return value of unapply;so it's a wrong syntax in the case of IsCompound has a boolean unapply
 * if: uu@IsCompound():unapply receive uu;
 */

五个if写明了所有的情况，代码可直接测试，测试时请对照if中的说明；

这里的unapply指的是IsCompound的unapply，

可以对它修改以不报错，但是原理就是这样的.更具体的解释如下：

val author1 = "aabc 111 1"
  val a Name()=author1
  println("a:"+a)
  val Name(b)=author1
  println("b:"+b)
  val c Name(d)=author1
  println("c:"+c+" "+"d:"+d)
  val h @Name(f,g)=author1
  println("h:"+h+" "+"f:"+f+" "+"g:"+g)
输出如下：
a:(aabc,111 1)
b:(aabc,111 1)
c:aabc d:111 1
h:aabc 111 1 f:aabc g:111 1

可见这与if所说的一致。

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