declare | dojo.declare optimized for node.js useage | Reflection library
kandi X-RAY | declare Summary
kandi X-RAY | declare Summary
declare is a node.js optimized version of dojo 1.7’s dojo.declare This example shows how to create a namespace and create classes within. No global namespace is violated. declare includes the folloing modified dojo functions. for more information check the dojo docs (with dojo. prefix for sure). var createNamespace = require("../index");. var myCustomNamespace = { "doesNotMatter" : "anyValue" };. var oop = createNamespace(myCustomNamespace);. var MYClass = oop.declare("my.namespace.MYCLASS",[],{ property:"value", constructor:function(){ console.log("I AM ALIVE"); }, methodA:function(){ console.log("methodA called") console.log(this); }, methodB:function(){ console.log("methodB called") } });. oop.declare("MYSubClass",[myCustomNamespace.my.namespace.MYCLASS],{ constructor:function(){ oop.mixin(this,{methodC:function(){ console.log("same mixin like known from dojo @" , this.declaredClass); }}); }, methodA:function(){ console.log("method A from " + this.declaredClass); } }). /* this will end up with the same result… / var instance = new MYClass(); / The namespace can be accessed with the object we used to create the "oop" instance. You could create more factories each with a different namespace */ var i2 = new myCustomNamespace.my.namespace.MYCLASS();. var subi = new myCustomNamespace.MYSubClass(); subi.methodA(); subi.methodB(); subi.methodC();.
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Top functions reviewed by kandi - BETA
- Declare class .
- find the current class .
- Recursively build a list of class names .
- Chain constructor .
- Make a constructor
- Make a new constructor
- Create a function that creates a class prototype chains .
- Mixin functions .
- Class instanceof .
- Apply a new instance to a constructor .
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Community Discussions
Trending Discussions on declare
QUESTION
Could you help me, I've got this error when I try building a project?
Oops! Something went wrong! :(
ESLint: 8.0.0
TypeError: Failed to load plugin '@typescript-eslint' declared in 'src.eslintrc': Class extends value undefined is not a constructor or null Referenced from: src.eslintrc
package.json
...ANSWER
Answered 2021-Oct-10 at 10:33https://github.com/typescript-eslint/typescript-eslint/issues/3982
It seems to be a compatibility problem
QUESTION
In my version of clang and libc++ (near HEAD
), this static_assert
passes:
ANSWER
Answered 2022-Mar-26 at 23:21std::vector
and other containers (except std::array
) are specified to have a copy constructor. This is not specified to be conditional on whether or not the element type is copyable. Only instantiation of the copy constructor's definition is forbidden if the element type is not copyable.
As a result std::is_copy_constructible_v
on the container will always be true
. There is no way to test whether an instantiation of a definition would be well-formed with a type trait.
It would be possible to specify that the copy constructor is not declared or excluded from overload resolution if the element type is not copyable. However, that would come with a trade-off which is explained in detail in this blog post: https://quuxplusone.github.io/blog/2020/02/05/vector-is-copyable-except-when-its-not/.
In short, if we want to be able to use the container with an incomplete type, e.g. recursively like
QUESTION
Today i have got this email:
Last July, we announced Advertising policy changes to help bolster security and privacy. We added new restrictions on identifiers used by apps that target children. When users choose to delete their advertising ID in order to opt out of personalization advertising, developers will receive a string of zeros instead of the identifier if they attempt to access the identifier. This behavior will extend to phones, tablets, and Android TV starting April 1, 2022. We also announced that you need to declare an AD_ID permission when you update your app targeting API level to 31 (Android 12). Today, we are sharing that we will give developers more time to ease the transition. We will require this permission declaration when your apps are able to target Android 13 instead of starting with Android 12.
Action Items If you use an advertising ID, you must declare the AD_ID Permission when your app targets Android 13 or above. Apps that don’t declare the permission will get a string of zeros. Note: You’ll be able to target Android 13 later this year. If your app uses an SDK that has declared the Ad ID permission, it will acquire the permission declaration through manifest merge. If your app’s target audience includes children, you must not transmit Android Advertising ID (AAID) from children or users of unknown age.
My app is not using the Advertising ID. Should i declare the AD_ID
Permission in Manifest or not?
ANSWER
Answered 2022-Mar-14 at 20:51Google describe here how to solve
https://support.google.com/googleplay/android-developer/answer/6048248?hl=en
Add in manifest
QUESTION
Today I updated my java version from 16 to 17 and I found that sealed
class is a new feature in it. I think it can be declared like this -
ANSWER
Answered 2021-Sep-17 at 15:04You can follow this link for examples.
In short sealed classes gives you the control of which models, classes etc. that can implement or extend that class/interface.
Example from the link:
QUESTION
class A {
int x;
static int i;
};
int x = 10;
int A::i = x;
...ANSWER
Answered 2022-Feb-07 at 10:07This is a peculiar language quirk - the scope resolution on the left, in int A::i
, affects the lookup scope on the right, so that actually refers to the x
member of A
.
Either rename one of the variables, or specify the scope of the desired x
explicitly:
QUESTION
I am aware of how ODR, linkage, static
, and extern "C"
work with functions. But I am not sure about visibility of types since they cannot be declared static
and there are no anonymous namespaces in C.
In particular, I would like to know the validity of the following code if compiled as C and C++
...ANSWER
Answered 2021-Oct-20 at 09:43For C. The program is valid. The only requirement that applies here is "strict aliasing rule" saying that the object can be accessed only via a l-value of a compatible type (+ a few exception described in 6.5p7).
The compatibility of structures/unions defined in separate translation units is defined in 6.2.7p1.
... two structure, union, or enumerated types declared in separate translation units are compatible if their tags and members satisfy the following requirements: If one is declared with a tag, the other shall be declared with the same tag. If both are completed anywhere within their respective translation units, then the following additional requirements apply: there shall be a one-to-one correspondence between their members such that each pair of corresponding members are declared with compatible types; if one member of the pair is declared with an alignment specifier, the other is declared with an equivalent alignment specifier; and if one member of the pair is declared with a name, the other is declared with the same name. For two structures, corresponding members shall be declared in the same order. For two structures or unions, corresponding bit-fields shall have the same widths. For two enumerations, corresponding members shall have the same values.
Therefore the structures are not compatible in the example.
However, it is not an issue because the f
object is created and accessed via locally defined type. UB would be invoked if the object was created with Foo
type defined in one translation unit and accessed via other Foo
type in the other translation unit:
QUESTION
If one defines a new variable in C++, then the name of the variable can be used in the initialization expression, for example:
...ANSWER
Answered 2021-Oct-06 at 22:12According to the C++17 standard (11.3.6 Default arguments)
9 A default argument is evaluated each time the function is called with no argument for the corresponding parameter. A parameter shall not appear as a potentially-evaluated expression in a default argument. Parameters of a function declared before a default argument are in scope and can hide namespace and class member name
It provides the following example:
QUESTION
Consider the following C++17 code:
...ANSWER
Answered 2021-Oct-03 at 12:09The code shown is valid (all C++ Standard versions, I believe). The similar restrictions are all listed in [reserved.names]. Since read
is not declared in the C++ standard library, nor in the C standard library, nor in older versions of the standard libraries, and is not otherwise listed there, it's fair game as a name in the global namespace.
So is it an implementation defect that it won't link with -static
? (Not a "compiler bug" - the compiler piece of the toolchain is fine, and there's nothing forbidding a warning on valid code.) It does at least work with default settings (though because of how the GNU linker doesn't mind duplicated symbols in an unused object of a dynamic library), and one could argue that's all that's needed for Standard compliance.
We also have at [intro.compliance]/8
A conforming implementation may have extensions (including additional library functions), provided they do not alter the behavior of any well-formed program. Implementations are required to diagnose programs that use such extensions that are ill-formed according to this International Standard. Having done so, however, they can compile and execute such programs.
We can consider POSIX functions such an extension. This is intentionally vague on when or how such extensions are enabled. The g++ driver of the GCC toolset links a number of libraries by default, and we can consider that as adding not only the availability of non-standard #include
headers but also adding additional translation units to the program. In theory, different arguments to the g++ driver might make it work without the underlying link step using libc.so
. But good luck - one could argue it's a problem that there's no simple way to link only names from the C++ and C standard libraries without including other unreserved names.
(Does not altering a well-formed program even mean that an implementation extension can't use non-reserved names for the additional libraries? I hope not, but I could see a strict reading implying that.)
So I haven't claimed a definitive answer to the question, but the practical situation is unlikely to change, and a Standard Defect Report would in my opinion be more nit-picking than a useful clarification.
QUESTION
I found the following macro in the source code of Perl:
...ANSWER
Answered 2021-Sep-03 at 09:35I guess it is used to make it an R-value, what makes it read-only in practice.
Example: One can write
QUESTION
#include
int main()
{
auto v = std::vector{std::vector{}};
return v.front().empty(); // error
}
...ANSWER
Answered 2021-Aug-30 at 07:46auto v = std::vector{std::vector{}};
actually creates a std::vector
because it uses std::vector
copy constructor. It is interpreted by the compiler as:
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