Nonescaping parameter type spelling

Using Swift
,
1

In the following fragment:

func foo(escaping: @escaping () -> (), nonescaping: () -> ()) {
    let a = escaping            // ok
    let b = nonescaping         // ok
    print(type(of: a))          // () -> ()
    print(type(of: b))          // () -> ()
    let c: () -> () = escaping  // ok
    let d: () -> () = nonescaping // 🛑 Using non-escaping parameter 'nonescaping' in a context expecting an @escaping closure
}

foo(escaping: {}, nonescaping: {})

both escaping and non escaping closure types are printed as () -> (), although this is apparently not the whole story and if I use this type in the explicit type position of a variable declaration the type doesn't match the nonescaping parameter. Is this type inexpressible in Swift?

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2

I think it is comparable to let vs. var:
No different type, but some restriction on what you can do with the variable/value.

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