private

Ada encapsulation and private types

由 蹲街弑〆低调 提交于 2021-01-28 12:12:11
问题 From the compiler's point of view, what's the difference between declaring an Ada type in a package spec or doing it inside the body? 回答1: Generally it is a good practice to make declarations (of types, but also other items like constants or subprograms) the most local possible. In your case if the type is used only in the body and not for users of your package specification (even as private type), put it in the body. Furthermore, if it is used only in a subprogram of the body, put it in that

Private inner module returning private item gives “private type in public interface” error

由 拥有回忆 提交于 2021-01-28 09:12:55
问题 In the below example, the module outer has a private type Private and a private inner module inner . inner is able to access Private (because child modules can access their parent's private items , even if they are not parked as public). inner defines a function not_really_public_interface() . While it is marked as public, it is really only available to outer because inner itself is not public. outer.rs struct Private; mod inner { use super::Private; pub fn not_really_public_interface() ->

Private inner module returning private item gives “private type in public interface” error

由 坚强是说给别人听的谎言 提交于 2021-01-28 09:12:03
问题 In the below example, the module outer has a private type Private and a private inner module inner . inner is able to access Private (because child modules can access their parent's private items , even if they are not parked as public). inner defines a function not_really_public_interface() . While it is marked as public, it is really only available to outer because inner itself is not public. outer.rs struct Private; mod inner { use super::Private; pub fn not_really_public_interface() ->

Generic to assess if a parameter is “instanceof a generic class” if constructor is private

由 最后都变了- 提交于 2021-01-05 09:11:40
问题 I am trying to assert in a function if an object is instance of a class with type predicate. function assertClass<CL extends new (...args: any[]) => any>(v: any, theClass: CL, message: string): v is CL { let is = v instanceof theClass ; assert(is, message); return is; } (credit) And it works except when the constructor of the class is private class Cat{} class Dog{ private constructor(){ } static async build(){ let ret = new Dog(); await somethingElse(); return ret; } } let animal: Cat | Dog;

Symfony2 access private services in tests

由 一个人想着一个人 提交于 2020-07-06 13:02:06
问题 Currently I'm working on testing some services in Symfony2 and I'm trying to use Guzzle MockPlugin for controlling CURL responses. Symfony version 2.3.8 is used. I've got to an interesting behaviour and I'm not sure if this is a Symfony2 bug or not. I have these services in services.yml: lookup_service_client: class: FOO public: false factory_service: lookup_client_builder factory_method: build lookup_repository_auth_type: class: AuthType arguments: ["@lookup_service_client"] lookup

C++中 public和private派生类继承问题和访问权限问题

由 浪尽此生 提交于 2020-04-22 03:40:08
本文链接地址：http://hi.baidu.com/laoyang1018/blog/item/933d5243a75c3f1a9313c6db.html ####################### //声明： 1 本帖作者是： 百变老羊 ，至此感谢！ 2 红色背景字体为个人阅读重点！ ####################### C++中 public和private派生类继承问题和访问权限问题 当一个子类从父类继承时，父类的所有成员成为子类的成员，此时对父类成员的访问状态由继承时使用的继承限定符决定。 1.如果子类从父类继承时使用的继承限定符是public，那么 (1)父类的public成员成为子类的public成员， 允许类以外的代码访问这些成员 ； (2)父类的private成员仍旧是父类的private成员，子类成员不可以访问这些成员； (3)父类的protected成员成为子类的protected成员，只允许子类成员访问； 2.如果子类从父类继承时使用的继承限定符是private，那么 (1)父类的public成员成为子类的private成员， 只允许子类成员访问 ； (2)父类的private成员仍旧是父类的private成员，子类成员不可以访问这些成员； (3)父类的protected成员成为子类的private成员，只允许子类成员访问； 3
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