Allow omission of generic type if its equal to the default type

Search Terms

generic, default, extends, equals, =

Suggestion

Allow the syntax extends = T in generics definition in place of extends T = T.

Currently

class Foo <T extends any = any> {}

Suggestion*

class Foo <T extends= any> {}

Use Cases

The problem of having to write <T extends any = any> revealed itself when I had a bit more complex type looking like keyof NetworkTransportProtocol that looked then somehow like this:

class Foo<T extends keyof system.NetworkTransportProtocol = keyof system.NetworkTransportProtocol> {}

which just was a lot of duplicated text.

class Foo<T extends= keyof system.NetworkTransportProtocol> {}

by reducing the duplicate information the whole syntax becomes better to read.

While I did focus on the default-case for class-generics. I believe this could also help with other definitions:

function foo <T extends=any>(bar: T) {}

Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

この issue では実装ファイルやテストが指定されていないため、まず TypeScript のジェネリック制約とデフォルト型に関する文法およびテストを見つけます。クラスと関数のジェネリックに対して提案された extends= 構文を比較し、続いて必要な parser と型チェックの挙動を判断します。提案された構文について、受け入れられた設計と、通過するテストカバレッジが揃えば完了です。