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.

The issue does not name implementation files or tests; start by locating TypeScript's grammar and tests for generic constraints and default types. Compare the proposed extends= syntax for class and function generics, then determine the parser and type-checking behavior required. Done would require an accepted design and passing coverage for the proposed syntax.