关于方法执行顺序的问题

scala中方法的优先级是根据方法名第一个字符决定的.scala

class Op(var x : Int) {
def +++(op: Op)  = { println(this.x+ " +++ "+op.x); this.x += op.x; this;} 
def ***(op: Op) = { println(this.x+ " *** " + op.x); this.x *= op.x ; this;}
}

val op1 = new Op(1)
val op2 = new Op(2)
val op3 = new Op(3)
val op4 = new Op(4)
op1 +++ op2 +++ op3 *** op4 //be equivalent to op1 +++ op2 +++ (op3 *** op4)

含有:方法.scala

class AA{
    var i= 0 
    def /: (s: String) = {
        println("s = " +s);
        s+"!"
    }

    def /: (aa : AA) = {
        println(aa.i);
        aa.i = aa.i+ 1
        this;
    }

    def is(s: String)= { println(s); i = i +1 ; this } 
    def isThis(s: String) = { println("isThis: "+s); i= i + 1; s}
    def add(i: Int) = { println("add "+i+" to "+this.i); this.i += i; this } 
}

val aa = new AA
aa is "tree" is "AA" /: aa // 等价于 aa.is("tree").is(aa./:("AA"))
aa is "tree" isThis "AA" /: aa // 等价于 aa.is("tree").isThis(aa./:("AA"))
aa is "tree" add 2 /: aa // 报错. (aa is "tree" add 2) /: aa 才等价于 aa./:(aa.is("tree").add(2))

含有方法.scala

class AA{
    var i= 0 
    def /: (s: String) = {
        println("s = " +s);
        s+"!"
    }

    def /: (aa : AA) = {
        println(aa.i);
        aa.i = aa.i+ 1
        this;
    }

    def is(s: String)= { println(s); i = i +1 ; this } 
    def isThis(s: String) = { println("isThis: "+s); i= i + 1; s}
    def add(i: Int) = { println("add "+i+" to "+this.i); this.i += i; this } 
}

val aa = new AA
aa is "tree" is "AA" /: aa // 等价于 aa.is("tree").is(aa./:("AA"))
aa is "tree" isThis "AA" /: aa // 等价于 aa.is("tree").isThis(aa./:("AA"))
aa is "tree" add 2 /: aa // 报错. (aa is "tree" add 2) /: aa 才等价于 aa./:(aa.is("tree").add(2))

注意执行的结果.
scala中，函数（或方法，运算符)的优先级是由函数名（或方法名，运算符）的第一个字符决定的。详细请查阅《Programming in Scala，2nd》第5.8节