Typescript 3.9.4:
type Test = ((value: { a: number }) => any) extends ((value: { [key: string]: any }) => any) ? true : false;
Based on the concept that {a: number} extends {[key: string]: any}, the expected type for Test should be true.
However, the actual type for Test turns out to be false, which raises some confusion. Is there a way to make this scenario work as intended?
This particular kind of type does not actually expand upon the other.
If we consider { a: number }, it is apparent that it indeed expands upon { [key: string]: number } when we think about calling a function with a signature like this:
foo(arg: { [key: string]: number })
In this case, it would be acceptable to provide an argument of type { a: number }. Any operation performed by foo using its argument could also apply to something of type { a: number }.
However, if you have:
foo(arg: (value: { [key: string]: number }) => any)
It would be inappropriate to pass an argument of type (value: { a: number }) => any, as the argument for foo requires a function capable of accepting an object with any string keys. Consider how foo might be implemented like this:
foo(arg: (value: { [key: string]: number }) => any) {
arg({ b: 7 });
}
In this scenario, the code would not be valid if arg has type (value: { a: number }) => any.
To delve deeper into this topic, what we are discussing here is known as type contravariance.
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