Haskell罕见的模式匹配

Question

In haskell, I can do:

s@"Hello" = s 

The result is:

>s
>"

It begins to print the String but never ends, what is goin on?

Answer 1

The @"Hello" is irrelevant here, all it does is fix the type to String . You get the same behaviour with

s :: String
s = s

Which is semantically equivalent to

s' :: String
s' = undefined

which gives the result

Prelude> s'
"*** Exception: Prelude.undefined

What I mean by “semantically equivalent” is, both s and s' are examples of bottom values , ie values from that “sin bin of error values any type contains, thanks to non-strictness”. As soon as you hit a bottom value, the pure Haskell language is basically powerless and has give in to, well, undefined , “impure behaviour”, like letting you wait forever or throwing an exception.

However, again thanks to non-strictness, this needs not necessarily happen. When printing a value, the first thing that happens is, the Show instance is invoked and asked to produce a string. A Haskell string is a lazy list. And the show of any string begins with " , therefore even if the string itself is utterly undefined, show will manage to produce that one character.

We can observe this more sheltered with

Prelude> head $ show s
'"'

Answer 2

In let s and top level expressions everything on the left-hand side of the = is in scope on the right-hand side. So you've created a looping "bottom" value.

Notice this behaves the same way:

Prelude> let s = (s::String)
Prelude> s
"

That's because (simplifying) print on String is defined something equivalent to:

printString chars = putChar '"' >> mapM_ putChar chars

Because chars is a loop the mapM_ putChar chars appears to hang.