Defining typescript generics with type safety

Question

Can you define generics with safe types, as you can with c#?

E.g.

public bool Foo<T>() where T : struct { /* */ }

Typescript now has generics, but can I perform a similar action?

Thanks.

Tim · Accepted Answer · 2013-07-15 13:56:04Z

104

Ok it seems you can do this:

Foo<T extends IBar>() { /* */ }

And that seems to make all calls require the T to implement IBar.

answered Jul 15, 2013 at 13:56

Tim

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+1 - spot on. This is taken from the Java implementation of generics - the extends keyword is used to constrain to either an interface or a class.

But it seems it does not enforce anything yet.

Note that if you want to constrain by interfaces, you use the extends keyword rather than the implements on. For example Foo<T extends IBar>.

Note that it's not required for a type to "implement" the IBar interface, it must merely be compatible with it.

Hypenate · Accepted Answer · 2021-04-10 12:11:05Z

0

If you don't want to create an extra class/interface, you can, for example, do the following:

Foo<T extends{ id: string | number }>

answered Apr 10, 2021 at 12:11

Hypenate

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