Kaedah overloading di JVM

Selamat datang ke blog Java Challengers baru ! Blog ini dikhaskan untuk konsep yang mencabar dalam pengaturcaraan Java. Kuasai mereka dan anda akan berjaya menjadi pengaturcara Java yang berkemahiran tinggi.

Teknik dalam blog ini memerlukan banyak usaha untuk dikuasai, tetapi mereka akan membuat perbezaan yang besar dalam pengalaman harian anda sebagai pembangun Java. Menghindari pepijat lebih mudah apabila anda mengetahui cara menerapkan teknik pengaturcaraan inti Java dengan betul, dan mengesan pepijat jauh lebih mudah apabila anda mengetahui dengan tepat apa yang berlaku pada kod Java anda.

Adakah anda bersedia untuk mula menguasai konsep teras dalam pengaturcaraan Java? Kemudian mari mulakan dengan Java Challenger pertama kami!  

Terminologi: Kaedah overloading

Kerana istilah overloading , pembangun cenderung menganggap teknik ini akan membebani sistem, tetapi itu tidak benar. Dalam pengaturcaraan, kaedah overloading bermaksud menggunakan nama kaedah yang sama dengan parameter yang berbeza. 

Apakah kaedah overloading?

Overloading kaedah adalah teknik pengaturcaraan yang membolehkan pembangun menggunakan nama kaedah yang sama berkali-kali dalam kelas yang sama, tetapi dengan parameter yang berbeza. Dalam kes ini, kami mengatakan bahawa kaedah ini berlebihan. Penyenaraian 1 menunjukkan satu kaedah yang parameternya berbeza dalam jumlah, jenis, dan urutan.

Penyenaraian 1. Tiga jenis kaedah overloading

 Number of parameters: public class Calculator { void calculate(int number1, int number2) { } void calculate(int number1, int number2, int number3) { } } Type of parameters: public class Calculator { void calculate(int number1, int number2) { } void calculate(double number1, double number2) { } } Order of parameters: public class Calculator { void calculate(double number1, int number2) { } void calculate(int number1, double number2) { } } 

Kaedah pemuatan dan jenis primitif

Dalam Penyenaraian 1, anda melihat jenis primitif intdan double. Kami akan bekerjasama dengan jenis ini dan lain-lain, jadi luangkan masa sebentar untuk mengkaji jenis primitif di Java.

Jadual 1. Jenis primitif di Jawa

Jenis Julat Lalai Saiz Contoh literal
 boolean  betul atau salah  salah  1 bit  betul salah
 byte  -128 .. 127  0  8 bit  1, -90, 128
 char  Watak unicode atau 0 hingga 65,536  \ u0000  16 bit  'a', '\ u0031', '\ 201', '\ n', 4
 short  -32,768 .. 32,767  0  16 bit  1, 3, 720, 22,000
 int  -2,147,483,648 .. 2,147,483,647  0  32 bit  -2, -1, 0, 1, 9
 long  -9,223,372,036,854,775,808 hingga 9,223,372,036,854,775,807  0  64 bit  -4000L, -900L, 10L, 700L
 float  3.40282347 x 1038, 1.40239846 x 10-45  0.0  32 bit  1.67e200f, -1.57e-207f, .9f, 10.4F
 double

 1.7976931348623157 x 10308, 4.9406564584124654 x 10-324

 0.0  64 bit  1.e700d, -123457e, 37e1d

Mengapa saya mesti menggunakan kaedah overloading?

Beban berlebihan menjadikan kod anda lebih bersih dan mudah dibaca, dan ini juga dapat membantu anda mengelakkan bug dalam program anda.

Berbeza dengan 1 Penyenaraian, bayangkan satu program di mana anda mempunyai pelbagai calculate()kaedah dengan nama-nama seperti calculate1, calculate2, calculate3. . . tidak bagus, bukan? Melebihi calculate()kaedah membolehkan anda menggunakan nama kaedah yang sama sambil hanya mengubah apa yang perlu diubah: parameternya. Sangat mudah untuk mencari kaedah yang terlalu banyak kerana mereka dikumpulkan dalam kod anda.

Apa yang berlebihan tidak

Ketahuilah bahawa menukar nama pemboleh ubah tidak berlebihan. Kod berikut tidak akan disusun:

 public class Calculator { void calculate(int firstNumber, int secondNumber){} void calculate(int secondNumber, int thirdNumber){} } 

Anda juga tidak boleh membebani kaedah dengan menukar jenis pengembalian dalam tandatangan kaedah. Kod berikut tidak akan disusun, sama ada:

 public class Calculator { double calculate(int number1, int number2){return 0.0;} long calculate(int number1, int number2){return 0;} } 

Lebihan konstruktor

Anda boleh memuatkan konstruktor dengan cara yang sama seperti kaedah:

 public class Calculator { private int number1; private int number2; public Calculator(int number1) {this.number1 = number1;} public Calculator(int number1, int number2) { this.number1 = number1; this.number2 = number2; } } 

Ikutilah kaedah mengatasi masalah!

Adakah anda bersedia untuk Java Challenger pertama anda? Mari kita ketahui!

Mulakan dengan menyemak kod berikut dengan teliti.

Penyenaraian 2. Cabaran kaedah muatan yang lebih maju

 public class AdvancedOverloadingChallenge3 { static String x = ""; public static void main(String... doYourBest) { executeAction(1); executeAction(1.0); executeAction(Double.valueOf("5")); executeAction(1L); System.out.println(x); } static void executeAction(int ... var) {x += "a"; } static void executeAction(Integer var) {x += "b"; } static void executeAction(Object var) {x += "c"; } static void executeAction(short var) {x += "d"; } static void executeAction(float var) {x += "e"; } static void executeAction(double var) {x += "f"; } } 

Baiklah, anda telah menyemak kodnya. Apakah outputnya?

  1. befe
  2. bfce
  3. kecekapan
  4. aecf

Lihat jawapan anda di sini.

Apa yang baru berlaku? Bagaimana JVM menyusun kaedah yang terlalu banyak

Untuk memahami apa yang berlaku dalam Penyenaraian 2, anda perlu mengetahui beberapa perkara mengenai bagaimana JVM menyusun kaedah yang terlalu banyak.

Pertama sekali, JVM secara bijak malas : ia akan sentiasa berusaha sedaya upaya untuk melaksanakan kaedah. Oleh itu, semasa anda memikirkan bagaimana JVM menangani kelebihan beban, ingatlah tiga teknik penyusun penting:

  1. Meluas
  2. Tinju (autoboxing dan unboxing)
  3. Varargs

Sekiranya anda tidak pernah menemui ketiga-tiga teknik ini, beberapa contoh akan membantu menjelaskannya. Perhatikan bahawa JVM melaksanakannya mengikut urutan yang diberikan .

Berikut adalah contoh pelebaran :

 int primitiveIntNumber = 5; double primitiveDoubleNumber = primitiveIntNumber ; 

Ini adalah urutan jenis primitif apabila dilebarkan:

Rafael del Nero

Berikut adalah contoh autoboxing :

 int primitiveIntNumber = 7; Integer wrapperIntegerNumber = primitiveIntNumber; 

Perhatikan apa yang berlaku di belakang tabir ketika kod ini disusun:

 Integer wrapperIntegerNumber = Integer.valueOf(primitiveIntNumber); 

Berikut adalah contoh  penyahpaparan :

 Integer wrapperIntegerNumber = 7; int primitiveIntNumber= wrapperIntegerNumber; 

Inilah yang berlaku di sebalik tabir ketika kod ini disusun:

 int primitiveIntNumber = wrapperIntegerNumber.intValue(); 

Dan berikut adalah contoh varargs ; perhatikan bahawa ia varargsadalah yang terakhir dilaksanakan:

 execute(int… numbers){} 

Apa itu varargs?

Used for variable arguments, varargs is basically an array of values specified by three dots (…) We can pass however many int numbers we want to this method.

For example:

execute(1,3,4,6,7,8,8,6,4,6,88...); // We could continue… 

Varargs is very handy because the values can be passed directly to the method. If we were using arrays, we would have to instantiate the array with the values.

Widening: A practical example

When we pass the number 1 directly to the executeAction method, the JVM automatically treats it as an int. That’s why the number doesn't go to the executeAction(short var) method.

Similarly, if we pass the number 1.0, the JVM automatically recognizes that number as a double.

Of course, the number 1.0 could also be a float, but the type is pre-defined. That’s why the executeAction(double var) method is invoked in Listing 2.

When we use the Double wrapper type, there are two possibilities: either the wrapper number could be unboxed to a primitive type, or it could be widened into an Object. (Remember that every class in Java extends the Object class.) In that case, the JVM chooses to wided the Double type to an Object because it takes less effort than unboxing would,  as I explained before.

The last number we pass is 1L, and because we've specified the variable type this time, it is long.

Video challenge! Debugging method overloading

Debugging is one of the easiest ways to fully absorb programming concepts while also improving your code. In this video you can follow along while I debug and explain the method overloading challenge:

Common mistakes with overloading

By now you’ve probably figured out that things can get tricky with method overloading, so let’s consider a few of the challenges you will likely encounter.

Autoboxing with wrappers

Java is a strongly typed programming language, and when we use autoboxing with wrappers there are some things we have to keep in mind. For one thing, the following code won't compile:

 int primitiveIntNumber = 7; Double wrapperNumber = primitiveIntNumber; 

Autoboxing will only work with the double type because what happens when you compile this code is the same as the following:

 Double number = Double.valueOf(primitiveIntNumber); 

The above code will compile. The first int type will be widened to double and then it will be boxed to Double. But when autoboxing, there is no type widening and the constructor from Double.valueOf will receive a double, not an int. In this case, autoboxing would only work if we applied a cast, like so:

 Double wrapperNumber = (double) primitiveIntNumber; 

Remember that Integer cannot be Long and Float cannot be Double. There is no inheritance. Each of these types--Integer, Long, Float, and Double--is a Number and an Object.

When in doubt, just remember that wrapper numbers can be widened to Number or Object. (There is a lot more to explore about wrappers but I will leave it for another post.)

Hard-coded number types in the JVM

When we don’t specify a type to a number, the JVM will do it for us. If we use the number 1 directly in the code, the JVM will create it as an int. If you try to pass 1 directly to a method that is receiving a short, it won’t compile.

For example:

 class Calculator { public static void main(String… args) { // This method invocation will not compile // Yes, 1 could be char, short, byte but the JVM creates it as an int calculate(1); } void calculate(short number) {} } 

The same rule will be applied when using the number 1.0; although it could be a float, the JVM will treat this number as a double:

 class Calculator { public static void main(String… args) { // This method invocation will not compile // Yes, 1 could be float but the JVM creates it as double calculate(1.0); } void calculate(float number) {} } 

Another common mistake is to think that the Double or any other wrapper type would be better suited to the method that is receiving a double. In fact, it takes less effort for the JVM to widen the Double wrapper to an Object instead of unboxing it to a double primitive type.

To sum up, when used directly in Java code, 1 will be int and 1.0 will be double. Widening is the laziest path to execution, boxing or unboxing comes next, and the last operation will always be varargs.

As a curious fact, did you know that the char type accepts numbers?

 char anyChar = 127; // Yes, this is strange but it compiles 

What to remember about overloading

Overloading is a very powerful technique for scenarios where you need the same method name with different parameters. It’s a useful technique because having the right name in your code makes a big difference for readability. Rather than duplicate the method and add clutter to your code, you may simply overload it. Doing this keeps your code clean and easy to read, and it reduces the risk that duplicate methods will break some part of the system.

What to keep in mind: When overloading a method the JVM will make the least effort possible; this is the order of the laziest path to execution:

  • First is widening
  • Second is boxing
  • Third is Varargs

What to watch out for: Tricky situations will arise from declaring a number directly: 1 will be int and 1.0 will be double.

Also remember that you can declare these types explicitly using the syntax of 1F or 1f for a float or 1D or 1d for a double.

That concludes our first Java Challenger, introducing the JVM’s role in method overloading. It is important to realize that the JVM is inherently lazy, and will always follow the laziest path to execution.

 

Answer key

The answer to the Java Challenger in Listing 2 is: Option 3. efce.

More about method overloading in Java

  • Java 101: Classes and objects in Java: A true beginner’s introduction to classes and objects, including short sections on methods and method overloading.
  • Java 101: Elementary Java language features: Learn more about why it matters that Java is a strongly typed language and get a full introduction to primitive types in Java.
  • Terlalu banyak parameter dalam metode Java, Bahagian 4: Terokai batasan dan kekurangan kaedah overloading, dan bagaimana mereka dapat diatasi dengan mengintegrasikan jenis dan objek parameter khusus.

Kisah ini, "Kaedah overloading di JVM" pada awalnya diterbitkan oleh JavaWorld.