scala函数

函数 必须给出参数类型，可是不必定给出函数的返回值类型，只要右侧函数体不包含递归语句，Scala就能够本身推断返回值类型
def 函数名(参数名:类型,参数名1:类型2):返回类型 = {}

explame:

def sayHello(name:String,age:Int) = { 
    print("name: "+name + " age: "+age)
}

:paste/** 多行函数用{}包含函数体 **/
def sayHello(name:String,age:Int) = { 
    if(age > 18){
        println("name: "+name + " age: "+age + " i am a audlt \n")
        age/** if是有返回值的 这里是age **/
    }else{
        println("name: "+name + " age: "+age + " i am a boy \n")
        age/** if是有返回值的 这里是age **/
    }
}

／** 单行函数 **／
scala> def sayHello(name:String) = println("hello " + name)
sayHello: (name: String)Unit

scala> sayHello("xp")
hello xp

/** 函数赋值给变量 **/
scala> def sayHello(name:String) = println("hello" + name)
sayHello: (name: String)Unit


scala> val sayHelloFun = sayHello _
sayHelloFun: String => Unit = <function1>

scala> sayHello
sayHello      sayHelloFun   

scala> sayHelloFun("xp")
helloxp

/** 匿名函数赋值给变量 **/
／** val 变量 = 参数列表 => 函数体 **／
scala> val sayHello = (name:String) => print("hello" + name)
sayHello: String => Unit = <function1>

scala> sayHello
sayHello      sayHelloFun   

scala> sayHello("DD")
helloDD

scala> val sayHello = (name:String) => print("hello" + name)
sayHello: String => Unit = <function1>

/** 高阶函数 **/
scala> def greeting(func:(String) => Unit,name:String) {func(name)}
greeting: (func: String => Unit, name: String)Unit

scala> greeting(sayHelloFun,"CC")
helloCC

/** 高阶函数返回函数 **/
scala> def greeting(name:String) = (name:String) => println("hello "+name)
greeting: (name: String)String => Unit

scala> val greet = greeting("hello")
greet: String => Unit = <function1>

scala> greet("xp")
hello xp