To declare a few variables and then assign the same meaning to them is not a problem. You can write, for example.int a, b, c, d; a = b = c = d = 0; However, if you want to initialize the variables when they are declared, each variable will have to be initiated separately. For exampleint a = 0, b = 0, c = 0, d = 0; orint a = 0, b = a, c = a, d = a; Simple announcement in C++ is defined as (7 Declarations)The simple-declaration attribute-specifier-seqopt decl-specifier-seqopt init-declarator-listopt; where init-declarator-list is defined as (8 Declarators)init-declarator-list:init-declarator init-declarator-list , init-declarator init-declarator:declarator initializeropt So every declarator (for simplicity: declared identifier) on the list has its own initiator.

The exact design parameter (as opposed to type) does not play a role. In the second leaflet, the value given in the parameter will not be recorded in the field of Employee. Add the field display challenge name After the creation of copies of the class in each case, for example, in the second case:public class TestEmployee { public static void main(String[] args) { String name = "larry"; Employee harry = new Employee(name); System.out.println(harry.name); } } class Employee { String name; public Employee(String name) { System.out.println("constructor"); } } In order to visually distinguish the name of the parameter and the name of the class field, the parameter was called nHe could also be named and nameBut then visually it would be hard to tell where. In any case, the compiler would have successfully dismantled such a design and co-compiled the class.The main point is that you must understand from this example: Just by calling the design parameter like the class field, you don't get automatic attribution.

As I know, Java, different from the C++ and lower languages, does not in itself initiate variables. ♪ ♪Let's say, "C++," int a; and not initiate it, after compiling and putting it on the screen, you can see that your variable was automatically triggered by a number of operational memory.But in Java, I don't know that... And if you try to use your uninitiated variable or object, the compiler will not compil your code, or eject NullPointerException... So to use it under conditions and other conditions, it's better to pre-empt this variable. Constant or 0 But you'll be there. 100% confident that this variable Never It won't take your constant, or it's 0, after which the variable was checked or not. ♪ ♪Like this:int a = -999; //code //Действия над переменной a //code if(a != -999){ System.out.print("Твоя переменная была изменена!"); }

How do you write the test initialized or not? and if not, initialize it.For example:public ArrayList<Part> initArrayList(ArrayList<Part> parts) { if (parts == null) { return new ArrayList<>(); } return parts; } UPD♪In view of the level of the question raised, he did not specifically want to confuse the author. http://sernam.ru/book_java.php?id=16 But if you know about him, the method. initArrayList(...) A few shorter can be achieved:public static ArrayList<Part> initArrayList(ArrayList<Part> parts) { return parts == null ? new ArrayList<>() : parts; } Actually, it is. syntactic sugarNot much.

temp2 located in the glass. When the designer comes out Cowsobject temp2 will be automatically destroyed = the constituent will produce discontents of all its signals.

Then it's like,class A: def __init__(self, **kwargs): self.__dict__.update(kwargs) kwargs - is the dictionary of arguments.Default option:class A: __defaults = dict(b1=1, b2='2', foo='bar') def __init__(self, **kwargs): self.__dict__.update(self.__defaults) self.__dict__.update(kwargs) Originally the class is initiated by default values (__defaultsand then the values from kwargs♪ If kwargs There are arguments that exist __defaultsnew values (from kwargsReplace the old ones. __defaults)Use the function https://fastcore.fast.ai/basics.html#store_attr Library https://pypi.org/project/fastcore/ , it takes all the parameters of the method and records their values in the object &apos; s attributions with the same names (if a parameter does not need to be specified in the attributes, it can be built, see documentation).from fastcore.utils import store_attr class B: def init(self, a, b1=1, b2=2, b3=333): store_attr() def update(self, a, b1, b2, b3): store_attr() b = B(a='aaa') print(b.a, b.b1, b.b2, b.b3) b.update('bbb', 3, 2, 1) print(b.a, b.b1, b.b2, b.b3)

How do the top, when the facility is set, assign the variable res meaning true?Put it in the designer, if this is the calculated value, make a normal announcement if it is a constant value:class Middle extends Top { private boolean res; public Middle(String path) { super(path); res = zvezdy_slozhilis(); } } class Middle extends Top { private boolean res = false; public Middle(String path) { super(path); } } Grade bottom extended middle contains methods and local variables to be operated by the new $top();I shouldn't. Class either has all the necessary contents in itself or declares certain methods abstract or uses copies of other classes, but I don't know anything about the heirs.♪

We can do this:public class MySocket{ private Socket mSocket; public MySocket(){ } public void setSocket(Socket socket){ mSocket = socket; } public Socket getSocket(){ return mSocket; } } And create a copy of it:final MySocket connectionSocket = new MySocket(); And then...try{ connectionSocket.setSocket(welcomeSocket.accept()); } catch (IOException e2) { e2.printStackTrace(); } And inside Runnable request connectionSocket.getSocket()♪

Because when you do thatMix4[] m4a = new Mix4[3] You only create a set of objects. If you see what's in this body, you'll see that there are three objects that refer to. null♪In order to fill the mass, it's exactly 16 lines from your code.The 12 row initiates the mass, 16 row initiates the elements of the mass.

It's just that we need to get this variable straight. Out of classclass A { public: static int count; A() { count++; // При создании каждого объекта эта переменная должна увеличиваться } }; //... int A::count = 0; You may also lower the trigger because the variable will in any case be initiated by zero:int A::count; If the definition of the class is placed in the headline file, the definition of the static variable should be placed in one of the programme modules.

There is the following quotation line from standard C+++, which explains how the elements of the mass are initiated when the mass is initialized.int frequency[frequencySize] = { 0 }; The first thing to draw attention is that the masses belong to Aggregate type♪ We therefore refer to the section where the initialization of the aggregate types 8.5.1 Aggregates is described:7 If there are fewer initializer-clauses in the list than there are members in the aggregate, then each member not explicitly initialized shall be initialized from its brace-or-equal-initializer or, if there is no brace-or-equalinitializer, from an empty initializer list(8.5.4).This extract from the standard indicates that if the initials are less than the elements in the body, the elements of the mass that do not have a clear initiator shall be initiated as if the empty figure brackets for each element indicated. I mean, you can imagine as follows:int frequency[frequencySize] = { 0, {}, {}, {}, {}, {}, {}, {}, {}, {}, {} }; We turn now to the standard for clarification, which means when there are empty figures. Section 8.5.4 List-initialization, paragraph 3, is written- Otherwise, if the initializer list has no elements, the object is value-initializedWhich means value-initialized? This is described in section 8.5 Initializers of the standard8 To value-initialize an object of type T means: - otherwise, the object is zero-initialized♪And finally, the notion zero-initialized means (8.5 Initializers) 6 To zero-initialize an object or reference of type T means: - if T is a scalar type (3.9), the object is initialized to the value obtained by converting the integer literal 0 (zero) to T;So all the elements of the mass that were not clearly inspired by the initiators are initiated by zero. That's why the record ended upint frequency[frequencySize] = { 0 }; equivalent to the following:int frequency[frequencySize] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }; And eventually, they get the tape.int frequency[frequencySize] = { 0 }; equivalentint frequency[frequencySize] = {}; Note that in C, the last two entries are not equivalent. For language C, recordint frequency[frequencySize] = {}; I'm wrong, and the compiler will give a misstatement.

findViewById for example onCreate() activate the method. Leave. public View customView; and onCreate add:customView = (View) findViewById(R.id.btn);UPD. If you dig deeper, Activity You're at the beginning of the method. onCreate() challenge. setContentView()♪ This is the challenge that triggers the marking. Activity and only after he's worked, findViewById()♪UPD2. Life cycle diagram ActivityThank you. http://developer.android.com/training/basics/activity-lifecycle/starting.html Thus, based on the diagram, initialization can be performed in: onCreate() after setContentView(), in methods onStart() and onResume()♪

In YiiBase there is a method:public static function createWebApplication($config=null){ return self::createApplication('CWebApplication',$config); } Let'spublic static function createApplication($class,$config=null) { return new $class($config); } The method createApplication will return a copy of Class CWebApplicationLet's go to the folder with the class:class CWebApplication extends CApplication{ ... } Let's go to CApplication.abstract class CApplication extends CModule { ... /** * Runs the application. * This method loads static application components. Derived classes usually overrides this * method to do more application-specific tasks. * Remember to call the parent implementation so that static application components are loaded. */ public function run() { ... } } That's the kind of run.

According to the https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html Fields that are declared but not initialized will be set to a reasonable default by the compiler.I mean, a compilator.

There you go.Terminal::Terminal(int x, int y):window(x, y) { } Take into account the procedure in which you (initiators) did not record them, the initialization procedure is basic designers, and then the members as they appear in the class announcement.

That's it.arr.i = { 4, 2, 1 }; Not initializationAnd attempt Appropriation♪ Initiation:struct array { int i[10]; } arr = { 4, 2, 1 }; Try that. ♪ ♪

Recently read out here that the same variable can be dynamically initiated after it was statically initiated. Is that correct? If so, can the variable be triggered twice?Yeah, that's right. Yeah, the variable can be initialized twice.It's rather a question of terminology: Is it right to call the second initialization... The language standard says that, at the beginning of the programme, static initialization of static storage variables with static storage class is carried out.static initialization) They may then be further exposed to dynamic initialization during the programme process (see annex).dynamic initialization)The fact of dual initialization could well be found in practice. http://eel.is/c++draft/basic.start.static#note-2 :#include <iostream> using std::cout; using std::endl; inline double fd() { return 1.0; } extern double d1; double d2 = d1; // unspecified: // may be statically initialized to 0.0 or // dynamically initialized to 0.0 if d1 is // dynamically initialized, or 1.0 otherwise double d1 = fd(); // may be initialized statically or dynamically to 1.0 int main() { cout << "d1: " << d1 << endl; cout << "d2: " << d2 << endl; } Conclusion of the programme http://coliru.stacked-crooked.com/a/4cddbbb257dc63b9 ♪ https://ideone.com/5Oc0F8 d1: 1 d2: 0 The observed conduct can be explained as follows.Removed d1 and d2 Statically initiated by zero.Then variable d2 dynamically initiated by variable value d1 (sighs)double d2 = d1;)Finally, variable d1 dynamically initiated by a function challenge fd (sighs)double d1 = fd();)Well, it's not that simple. If not the variable d2codeinline double fd() { return 1.0; } double d1 = fd(); in general is equivalent to that code:double d1 = 1.0; Dynamic initiation of variables d1 can be replaced by static initialization, with variable value d1 No change. The language standard clearly permits such a replacement. See: http://eel.is/c++draft/basic.start.static#3 ♪ Stage of static initialization of variable d1 may be assigned value 1.0Not 0.0♪ That is, the observed may be such as:Modified d1 Statically initiated by the value 1.0and variable d2 Statically initiated by the value 0.0♪Then variable d2 dynamically initiated by variable value d1 (i.e. also takes importance 1.0)Finally, as a variable d2 eventually very specific, known at the compilation stage, it, as well as the variable d1may not be subject to dynamic initialization, but only static.Something like that.

The figurative brackets in the initiation of the variable are based on its type. And if there are very many brackets attached to each other, the automatic definition of types does not work. We need to tell the compiler what type you mean.# include <string> # include <array> constexpr int A = 2; constexpr int B = 5; struct prasetoReplace { std::wstring find; std::wstring replace; }; extern const std::array <std::array<prasetoReplace, A>,B> aaa; const std::array <std::array<prasetoReplace, A>, B> aaa = { std::array<prasetoReplace, A> { prasetoReplace {L"1 ракушек", L"1 ракушку"},{L"1 ракушек", L"1 ракушку"}}, { prasetoReplace {L"1 ракушек", L"1 ракушку"},{L"1 ракушек", L"1 ракушку"}}, { prasetoReplace {L"1 ракушек", L"1 ракушку"},{L"1 ракушек", L"1 ракушку"}}, { prasetoReplace {L"1 ракушек", L"1 ракушку"},{L"1 ракушек", L"1 ракушку"}}, { prasetoReplace {L"1 ракушек", L"1 ракушку"},{L"1 ракушек", L"1 ракушку"}}}; To understand the problem: this is the minimum example of the ambiguity of the design.class A{ public: A(int x):i{x}{} int i; }; class B{ public: B(int x):i{x}{} int i; }; class C{ public: C(A a):i{a.i}{} C(B b):i{b.i}{} int i; }; // error: call of overloaded 'C(<brace-enclosed initializer list>)' is ambiguous // вызов конструктора неоднозначен : или A или B // C c0 { { 0 } } ; C c1 { A { 1 } } ; C c2 { B { 2 } } ;

The default designer is a designer with no parameters (or parameters are default). The designer with two parameters in your case is not a default designer.What's going on with the tape Point a{1, 2}; Name http://eel.is/c++draft/dcl.init.aggr ♪ It is available in cases where there is no user design at the type. You can read it in more detail. https://ru.stackoverflow.com/a/495385/176217 ♪

In the first case, the facility is established and immediately removed. Since he's got a side effect designer, it could have some meaning (e.g., it could be a class that puts it in a file), so the compiler's gonna be honest.Look at the second case. https://ru.stackoverflow.com/a/1271908/177993 @user7860670