enum | Strongly typed enums in PHP supporting autocompletion | SDK library
kandi X-RAY | enum Summary
kandi X-RAY | enum Summary
This package offers strongly typed enums in PHP. In this package, enums are always objects, never constant values on their own. This allows for proper static analysis and refactoring in IDEs.
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Top functions reviewed by kandi - BETA
- Asserts that two values are equal .
- Asserts that the given value is an enum value .
- Assert value is an enum label .
- Returns a random enum .
- Get random value .
- Get random element label
- Convert value to Enum instance .
- Create an exception for a method name .
enum Key Features
enum Examples and Code Snippets
def _convert_tflite_enum_type_to_tf_type(tflite_enum_type):
"""Converts tflite enum type (eg: 0) to tf type (eg: tf.float32).
Args:
tflite_enum_type: tflite enum type (eg: 0, that corresponds to float32)
Raises:
ValueError: If an inva
def _convert_external_state_policy_to_enum(external_state_policy):
if isinstance(external_state_policy, options_lib.ExternalStatePolicy):
return external_state_policy
if external_state_policy == "warn":
return options_lib.ExternalStatePol
Community Discussions
Trending Discussions on enum
QUESTION
I stumbled upon a strange compile error in Clang-12. The code below compiles just fine in GCC 9. Is this a bug in the compiler or is there an actual problem with my code and GCC is just too forgiving?
...ANSWER
Answered 2022-Mar-19 at 11:22Looks like a bug to me. Modifying case A
to case nullptr
gives the following error message (on the template definition):
QUESTION
Consider this example
...ANSWER
Answered 2022-Feb-22 at 07:03Yes, as of now, the One Definition Rule in the C++ standard doesn't include enumerators.
However, the "the second a
is a redeclaration of the first a
" explanation doesn't work too.
From [dcl.enum#nt:enumerator-list] we can know that an enumerator-list is a list of enumerator-definition, so they're all definitions.
QUESTION
A simple question: is enum { a } e = 1;
valid?
In other words: does assigning a value, which isn't present in the set of values of enumeration constants, lead to well-defined behavior?
Demo:
...ANSWER
Answered 2022-Feb-05 at 14:09From the C18 standard in 6.7.2.2:
Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined, but shall be capable of representing the values of all the members of the enumeration.
So yes enum { a } e = 1;
is valid. e
is a 'integer' type so it can take the value 1
. The fact that 1
is not present as an enumeration value is no issue.
The enumeration members only give handy identifiers for some of possible values.
QUESTION
My .eslintrc.json
is:
ANSWER
Answered 2022-Jan-11 at 17:06It looks like you have defined custom paths in your TypeScript config (usually tsconfig.json
). The import
plugin doesn't know about the correct location of the TypeScript config and hence cannot resolve those paths. What you need to do, is to specify the correct path to your TypeScript config via the project
parameter in the resolver options:
QUESTION
Okay, I wanted to know why my preview wasn't working after updating Xcode. So I have an enum that looks like this.
...ANSWER
Answered 2021-Dec-21 at 14:58Please see my related answer I posted a couple of months ago on the Apple Developer Forums located at: https://developers.apple.com/forums/thread/681571?answerId=690251022#690251022 . Does this work for you?
QUESTION
My Code:
...ANSWER
Answered 2022-Jan-12 at 07:13Open Visual Studio Code settings and ensure the intelephense.environment.phpVersion
directive is set to "8.1.0"
. It can be set for the whole program or for specific projects.
QUESTION
I am not using AWS AppSync for this app. I have created Graphql schema, I have made my own resolvers. For each create, query, I have made each Lambda functions. I used DynamoDB Single table concept and it's Global secondary indexes.
It was ok for me, to create an Book item. In DynamoDB, the table looks like this: .
I am having issue with the return Graphql queries. After getting the Items
from DynamoDB table, I have to use Map function then return the Items
based on Graphql type
. I feel like this is not efficient way to do that. Idk the best way query data. Also I am getting null both author and authors query.
This is my gitlab-branch.
This is my Graphql Schema
...ANSWER
Answered 2022-Jan-09 at 17:06TL;DR You are missing some resolvers. Your query resolvers are trying to do the job of the missing resolvers. Your resolvers must return data in the right shape.
In other words, your problems are with configuring Apollo Server's resolvers. Nothing Lambda-specific, as far as I can tell.
Write and register the missing resolvers.GraphQL doesn't know how to "resolve" an author's books, for instance. Add a Author {books(parent)}
entry to Apollo Server's resolver map. The corresponding resolver function should return a list of book objects (i.e. [Books]
), as your schema requires. Apollo's docs have a similar example you can adapt.
Here's a refactored author
query, commented with the resolvers that will be called:
QUESTION
For example, I have Subscriber
class and its objects subscriber1
, subscriber2
, subscriber3
.
Is it possible to create an enum
that contains these objects?
I'm trying to do this
ANSWER
Answered 2021-Dec-16 at 18:44The value in an enum class are always of the type of the enum. Though you can also add fields and methods to your enum class:
QUESTION
I want to have a simple enumDescr
function for any Scala 3 enum.
Example:
...ANSWER
Answered 2021-Oct-27 at 23:45I don't see any common trait shared by all enum
companion objects.
You still can invoke the values
reflectively, though:
QUESTION
I am getting an error message with Xcode 13 on the following code
...ANSWER
Answered 2021-Aug-12 at 08:49See this Swift bug for an explanation and workaround from the Swift compiler team.
This is intentional. The ABI of enum cases with payloads that are potentially unavailable is not well-defined. That this worked in the past was by coincidence of your application not requiring the (potentially unavailable) type metadata for the payload. Please either increase your deployment target or mark Foo itself as available as the least-available case. https://github.com/apple/swift/pull/36327
So you either need to mark the whole enum as @available(iOS 13.0, *)
or need to increase your deployment target to iOS 13.0
.
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