參數化陣列(Array)類型

Parameterize Arrays




內容

  1. 1 簡要的語法範例
    1. 1.1 Array表示符 ( )
    2. 1.2 For Java Developer
      1. 1.2.1 相同處
      2. 1.2.2 不同處
    3. 1.3 Examples
  2. 2 Parameterize Arrays with types



簡要的語法範例


scala> val arrayDemo: Array[Int] = new Array(5)
arrayDemo: Array[Int] = Array(0, 0, 0, 0, 0)

scala> arrayDemo
res2: Array[Int] = Array(0, 0, 0, 0, 0)


Array表示符 ( )

So the first element in a Scala array named steps is steps(0), not steps[0].


For Java Developer


相同處

  1. 實例化同樣使用 new 關鍵字
  2. 相同類型的元素值可變序列, Array是一個元素值可變的物件
  3. 實例化後長度不可變

不同處

  1. 宣告 Array 不加 [ ]
  2. 實例化 Array 語句為 new Array[Type](length)
  3. 存取 Array 索引使用 (index)
  4. Scala 的 Array 優於 Java 可使用泛型化參數與類型變數
    1. Array[String](10)
    2. Array[T](10)
  5. Scala陣列是一個所有object都共享相同類型的可變序列。


Examples


Scala 的快速排序, http://www.qqread.com/java/2009/08/f472573.html




Parameterize Arrays with types

In addition to being functional, Scala is object-oriented. In Scala, as in Java, you define a blueprint for objects with classes. From a class blueprint, you can instantiate objects, or class instances, by using new. For example, the following Scala code instantiates a new String and prints it out: 

val s = new String("Hello, world!")
println(s)

In the previous example, you parameterize the String instance with the initial value "Hello, world!". You can think of parameterization as meaning configuring an instance at the point in your program that you create that instance. You configure an instance with values by passing objects to a constructor of the instance in parentheses, just like you do when you create an instance in Java. If you place the previous code in a new file named paramwithvalues.scala and run it with scala paramswithvalues.scala, you'll see the familiar Hello, world! greeting printed out.

In addition to parameterizing instances with values at the point of instantiation, you can in Scala also parameterize them with types. This kind of parameterization is akin to specifying a type in angle brackets when instantiating a generic type in Java 5 and beyond. The main difference is that instead of the angle brackets used for this purpose in Java, in Scala you use square brackets. Here's an example: 

val greetStrings = new Array[String](3)

greetStrings(0) = "Hello"
greetStrings(1) = ", "
greetStrings(2) = "world!\n"

for (i <- 0 to 2)
  print(greetStrings(i))

In this example, greetStrings is a value of type Array[String] (say this as, “an array of string”) that is initialized to length 3 by passing the value 3 to a constructor in parentheses in the first line of code. Type this code into a new file called paramwithtypes.scala and execute it with scala paramwithtypes.scala, and you'll see yet another Hello, world! greeting. Note that when you parameterize an instance with both a type and a value, the type comes first in its square brackets, followed by the value in parentheses.

Had you been in a more explicit mood, you could have specified the type of greetStrings explicitly like this: 

val greetStrings: Array[String] = new Array[String](3)
// ...

Given Scala's type inference, this line of code is semantically equivalent to the actual first line of code in paramwithtypes.scala. But this form demonstrates that while the type parameterization portion (the type names in square brackets) form part of the type of the instance, the value parameterization part (the values in parentheses) do not. The type of greetStrings is Array[String], not Array[String](3).

The next three lines of code in paramwithtypes.scala initializes each element of the greetStrings array: 

// ...
greetStrings(0) = "Hello"
greetStrings(1) = ", "
greetStrings(2) = "world!\n"
// ...

As mentioned previously, arrays in Scala are accessed by placing the index inside parentheses, not square brackets as in Java. Thus the zeroeth element of the array is greetStrings(0), not greetStrings[0] as in Java.

These three lines of code illustrate an important concept to understand about Scala concerning the meaning of val. When you define a variable with val, the variable can't be reassigned, but the object to which it refers could potentially still be mutated. So in this case, you couldn't reassign greetStrings to a different array; greetStrings will always point to the same Array[String] instance with which it was initialized. But you can change the elements of that Array[String] over time, so the array itself is mutable.

The final two lines in paramwithtypes.scala contain a for comprehension that prints out each greetStrings array element in turn. 

// ...
for (i <- 0 to 2)
  print(greetStrings(i))

The first line of code in this for comprehension illustrates another general rule of Scala: if a method takes only one parameter, you can call it without a dot or parentheses. to is actually a method that takes one Int argument. The code 0 to 2 is transformed into the method call 0.to(2). (This to method actually returns not an Array but a Scala iterator that returns the values 0, 1, and 2.) Scala doesn't technically have operator overloading, because it doesn't actually have operators in the traditional sense. Characters such as +, -, *, and /, have no special meaning in Scala, but they can be used in method names. Thus, the expression 1 + 2, which was the first Scala code you typed into the interpreter in Step 1, is essential in meaning to 1.+(2), where + is the name of a method defined in class scala.Int.

Another important idea illustrated by this example will give you insight into why arrays are accessed with parentheses in Scala. Scala has fewer special cases than Java. Arrays are simply instances of classes like any other class in Scala. When you apply parentheses to a variable and pass in some arguments, Scala will transform that into an invocation of a method named apply. So greetStrings(i) gets transformed into greetStrings.apply(i). Thus accessing the element of an array in Scala is simply a method call like any other method call. What's more, the compiler will transform any application of parentheses with some arguments on any type into an apply method call, not just arrays. Of course it will compile only if that type actually defines an apply method. So it's not a special case; it's a general rule.

Similarly, when an assignment is made to a variable that is followed by some arguments in parentheses, the compiler will transform that into an invocation of an update method that takes two parameters. For example, 

greetStrings(0) = "Hello"

will essentially be transformed into 

greetStrings.update(0, "Hello")

Thus, the following Scala code is semantically equivalent to the code you typed into paramwithtypes.scala: 

val greetStrings = new Array[String](3)

greetStrings.update(0, "Hello")
greetStrings.update(1, ", ")
greetStrings.update(2, "world!\n")

for (i <- 0.to(2))
  print(greetStrings.apply(i))

Scala achieves a conceptual simplicity by treating everything, from arrays to expressions, as objects with methods. You as the programmer don't have to remember lots of special cases, such as the differences in Java between primitive and their corresponding wrapper types, or between arrays and regular objects. However, it is significant to note that in Scala this uniformity does not usually come with a performance cost as it often has in other languages that have aimed to be pure in their object orientation. The Scala compiler uses Java arrays, primitive types, and native arithmetic where possible in the compiled code. Thus Scala really does give you the best of both worlds in this sense: the conceptual simplicity of a pure object-oriented language with the runtime performance characteristics of language that has special cases for performance reasons.


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