Scala学习笔记02_函数入门

in #cn7 years ago

函数入门

函数的定义与调用,在Scala中定义函数时,需要定义函数的函数名、参数、函数体。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sayHello(name:String, age:Int) = {
  if(age >= 18) {
    printf("Hi, %s, you are a big boy!\n", name)
    age
  } else {
      printf("Hi, %s, you are a children!\n", name)
      age
    }
}

// Exiting paste mode, now interpreting.

sayHello: (name: String, age: Int)Int

scala> sayHello("padluo", 30)
Hi, padluo, you are a big boy!
res1: Int = 30

Scala要求必须给出所有参数的类型,但是不一定给出函数返回值的类型,只要右侧的函数体不包含递归的语句,Scala就可以自己根据右侧的表达式推断出返回类型。

在代码块中定义包含多行语句的函数体

单行的函数,

scala> def sayHello(name:String) = printf("Hello, " + name)
sayHello: (name: String)Unit

scala> sayHello("padluo")
Hello, padluo

如果函数体中有多行代码,则可以用代码块的方式包裹多行代码,代码块中最后一行的返回值就是整个函数的返回值,与Java中不同,不是使用return返回值的。

# 函数定义没有写=
scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum(n:Int) {
  var result = 0
  for(i <- 1 to n) {
    result += i
  }
}

// Exiting paste mode, now interpreting.

sum: (n: Int)Unit

# 函数定义写了=,但是最后一行是赋值语句
scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum(n:Int) = {
  var result = 0
  for(i <- 1 to n) {
    result += i
  }
}

// Exiting paste mode, now interpreting.

sum: (n: Int)Unit

# 正确定义
scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum(n:Int) = {
  var result = 0
  for(i <- 1 to n) {
    result += i
  }
  result
}

// Exiting paste mode, now interpreting.

sum: (n: Int)Int

递归函数与返回类型,如果在函数体内递归调用函数自身,则必须手动给出函数的返回类型。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def fab(n:Int):Int = {
  if(n<=0) 1
  else fab(n-1) + fab(n-2)
}

// Exiting paste mode, now interpreting.

fab: (n: Int)Int

scala> fab(10)
res5: Int = 144

默认参数和带名参数

默认参数,在Scala中,有时我们调用某些函数时,不希望给出具体的参数值,而希望使用参数自身默认的值,此时就在定义函数时使用默认参数。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sayHello(name:String, age:Int=20) {
  print("Hello, " + name + ", your age is " + age)
}

// Exiting paste mode, now interpreting.

sayHello: (name: String, age: Int)Unit

scala> sayHello("leo")
Hello, leo, your age is 20
scala> sayHello("leo", 30)
Hello, leo, your age is 30

如果给出的参数不够,则会从左到右依次应用参数。

带名参数,在调用函数时,也可以不按照函数定义的参数顺序来传递参数,而是使用带名参数的方式来传递。

scala> sayHello(age=30, name="leo")
Hello, leo, your age is 30

还可以混合使用未命名参数和带名参数,但是未命名参数必须排在带名参数前面。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sayHello(firstName:String, middleName:String="Willian",lastName:String="Croft") = print(firstName + " " + middleName + " " + lastName)

// Exiting paste mode, now interpreting.

sayHello: (firstName: String, middleName: String, lastName: String)Unit

scala> sayHello("leo")
leo Willian Croft
scala> sayHello("leo", lastName="Jack", middleName="Tom")
leo Tom Jack

变长参数

如果函数体包含在花括号中,但没有前面的=号,那么返回类型就是Unit,这样的函数被称为过程,过程不返回值,我们调用它仅仅是为了它的副作用,有人不喜欢这种简明写法定义过程,并建议大家总是显式声明Unit返回类型。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum(nums:Int*) {
  var result = 0
  for(num <- nums) {
    result += num
  }
  result
}

// Exiting paste mode, now interpreting.

sum: (nums: Int*)Unit

scala> sum(1,2,3,4,5)

scala>

在Scala中,有时我们需要将函数定义为参数个数可变的形式,则此时可以使用变长参数定义函数。

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum(nums:Int*) = {
  var result = 0
  for(num <- nums) {
    result += num
  }
  result
}

// Exiting paste mode, now interpreting.

sum: (nums: Int*)Int

scala> sum(1,2,3,4,5)
res0: Int = 15

使用序列调用变长参数,如果想要将一个已有的序列直接调用变长参数函数,是不对的,如val s = sum(1 to 5)。此时需要使用Scala特殊的语法将参数定义为序列sum(1 to 5: _*),让Scala解释器能够识别。

scala> 1 to 5
res6: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5)

scala> sum(1 to 5)
<console>:13: error: type mismatch;
 found   : scala.collection.immutable.Range.Inclusive
 required: Int
       sum(1 to 5)

scala> sum(1 to 5:_*)
res3: Int = 15

scala> sum(1 to 5: _*)
res4: Int = 15

scala> sum(1 to 5 : _*)
res5: Int = 15

使用递归函数实现累加,

scala> :paste
// Entering paste mode (ctrl-D to finish)

def sum2(nums:Int*): Int = {
  if (nums.length == 0) 0
  else nums.head + sum2(nums.tail: _*)
}

// Exiting paste mode, now interpreting.

sum2: (nums: Int*)Int

scala> sum(1,2,3,4,5)
res8: Int = 15

过程、lazy值和异常

过程,在Scala中,定义函数时,如果函数体直接包裹在了花括号里面,而没有使用=连接,则函数的返回值类型就是Unit。这样的函数称为过程。过程通常用于不需要返回值的函数。

过程还有一种写法,就是将函数的返回值类型定义为Unit。

# 函数,没有显式声明返回值类型,自动推断返回值类型
scala> def sayHello(name:String) = "Hello, " + name
sayHello: (name: String)String

scala> sayHello("padluo")
res1: String = Hello, padluo

# 过程,没有使用=连接,调用它仅仅是为了它的副作用(print ...),即使块最后的表达式有值,整个函数最终是没有值返回的
scala> def sayHello(name:String) {print("Hello, " + name); "Hello, " + name}
sayHello: (name: String)Unit

scala> sayHello("padluo")
Hello, padluo

# 显式声明返回值类型为Unit,即使块最后的表达式有值,函数最终也是没有值返回
scala> def sayHello(name:String): Unit = "Hello," + name
sayHello: (name: String)Unit

scala> sayHello("padluo")

scala>

lazy值,如果将一个变量声明为lazy,则只有在第一次使用该变量时,变量对应的表达式才会发生计算,这种特性对于特别耗时的操作特别有用,比如打开文件进行IO,进行网络IO等。

scala> import scala.io.Source._
import scala.io.Source._

scala> lazy val lines = fromFile("/home/hadoop/test.txt").mkString
lines: String = <lazy>

scala> print(lines)
Hello World

scala> lazy val lines = fromFile("/home/hadoop/test1.txt").mkString
lines: String = <lazy>

scala> print(lines)
java.io.FileNotFoundException: /home/hadoop/test1.txt (No such file or directory)
  at java.io.FileInputStream.open(Native Method)
  at java.io.FileInputStream.<init>(FileInputStream.java:146)
  at scala.io.Source$.fromFile(Source.scala:91)
  at scala.io.Source$.fromFile(Source.scala:76)
  at scala.io.Source$.fromFile(Source.scala:54)
  at .lines$lzycompute(<console>:14)
  at .lines(<console>:14)
  ... 32 elided
  
scala> val lines = fromFile("/home/hadoop/test1.txt").mkString
java.io.FileNotFoundException: /home/hadoop/test1.txt (No such file or directory)
  at java.io.FileInputStream.open(Native Method)
  at java.io.FileInputStream.<init>(FileInputStream.java:146)
  at scala.io.Source$.fromFile(Source.scala:91)
  at scala.io.Source$.fromFile(Source.scala:76)
  at scala.io.Source$.fromFile(Source.scala:54)
  ... 32 elided

scala> def getLines = fromFile("/home/hadoop/test1.txt").mkString
getLines: String

scala> getLines
java.io.FileNotFoundException: /home/hadoop/test1.txt (No such file or directory)
  at java.io.FileInputStream.open(Native Method)
  at java.io.FileInputStream.<init>(FileInputStream.java:146)
  at scala.io.Source$.fromFile(Source.scala:91)
  at scala.io.Source$.fromFile(Source.scala:76)
  at scala.io.Source$.fromFile(Source.scala:54)
  at .getLines(<console>:14)
  ... 32 elided

val lines = fromFile("/home/hadoop/test1.txt").mkString,即使文件不存在也不会报错,只有第一个使用变量时会报错,证明了表达式计算的lazy特性。

异常

scala> :paste
// Entering paste mode (ctrl-D to finish)

try {
  throw new IllegalArgumentException("illegal argument!")
} catch {
  case _: IllegalArgumentException => print("sorry, error!")
} finally {
  print("\nrelease io resources!")
}

// Exiting paste mode, now interpreting.

sorry, error!
release io resources!
scala> import java.io._
import java.io._

scala> :paste
// Entering paste mode (ctrl-D to finish)

try {
  throw new IOException("user defined exception")
} catch {
  case e1: IllegalArgumentException => println("illegal argument")
  case e2: IOException => print("io exception")
}

// Exiting paste mode, now interpreting.

io exception

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