问题
I have a function a that should returns any if no generic type is provided, T otherwise.
var a = function<T>() : T {
return null;
}
var b = a<number>(); //number
var c = a(); //c is {}. Not what I want... I want c to be any.
var d; //any
var e = a<typeof d>(); //any
Is it possible? (Without changing the function calls obviously. AKA without a<any>().)
回答1:
Is it possible? (Without changing the function calls obviously. AKA without a().)
Yes.
I believe in your case you would do
var a = function<T = any>() : T {
return null;
}
Generic defaults were introduced in TS 2.3.
Default types for generic type parameters have the following syntax:
TypeParameter :
BindingIdentifier Constraint? DefaultType?
DefaultType :
`=` Type
For example:
class Generic<T = string> {
private readonly list: T[] = []
add(t: T) {
this.list.push(t)
}
log() {
console.log(this.list)
}
}
const generic = new Generic()
generic.add('hello world') // Works
generic.add(4) // Error: Argument of type '4' is not assignable to parameter of type 'string'
generic.add({t: 33}) // Error: Argument of type '{ t: number; }' is not assignable to parameter of type 'string'
generic.log()
const genericAny = new Generic<any>()
// All of the following compile successfully
genericAny.add('hello world')
genericAny.add(4)
genericAny.add({t: 33})
genericAny.log()
See https://github.com/Microsoft/TypeScript/wiki/Roadmap#23-april-2017 and https://github.com/Microsoft/TypeScript/pull/13487
回答2:
Is it possible? (Without changing the function calls obviously. AKA without a().)
No.
PS
Note that having a generic type that isn't actively used in any function parameters is almost always a programming error. This is because the following two are equavalent:
foo<any>() and <someEquvalentAssertion>foo() and leaves it completely at the mercy of the caller.
PS PS
There is an official issue requesting this feature : https://github.com/Microsoft/TypeScript/issues/2175
来源:https://stackoverflow.com/questions/34938617/typescript-force-default-generic-type-to-be-any-instead-of