Created
November 25, 2011 06:11
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关于方法执行顺序的问题
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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) |
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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)) |
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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)) |
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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