swift Parameter is implicitly non-escaping

Question

func test(f: () -> ()) {
    let a = f
    let b: () -> () = f // error
    
    a()
    b()
}

The only difference between a and b is whether the type is specified or not.

But what is the reason for the error to be printed only in b?

Please explain the clear difference between a and b.

user_6950415 · Answer

The error is shown because 'b' and 'f' do have different signatures.

'b' is an escaping closure 'f' is not.

when doing:

let a = f

you are copying 'f'. The signature stays the same "nonescaping () -> ()".

Declaring a closure in function definition gives you implicit 'nonescaping' in body gives you implicit 'escaping'.

Decorating your closure in the header with @escaping will solve the problem.

func test(f: @escaping () -> ()) {