问题
I understand that in:
f x = x + 1 where !y = undefined
the meaning of the bang pattern is that y is to be evaluated before f.
Similarly:
f x = x + 1 where !(!a, !b) = (undefined, undefined)
the meaning is the same, w.r.t x and y.
But what do the bang patterns mean in:
f x = x + 1 where (!a, !b) = (undefined, undefined)
It doesn't seem to cause undefined to be evaluated. When do in-tuple bang patterns come into effect? If the pattern's tuple is forced? Can anyone give an example where (!a, !b) = (..) differs from (a, b) = (..)?
回答1:
A bang pattern on the tuple itself will force evaluation of the tuple but not its elements. Bang patterns on the tuple elements will force them whenever the tuple itself is evaluated.
Here's an example of the differing behavior:
Prelude> let x = a + 1 where (a, b) = (1, undefined)
Prelude> x
2
Prelude> let x = a + 1 where (!a, !b) = (1, undefined)
Prelude> x
*** Exception: Prelude.undefined
回答2:
If you translate it to let:
f x = let (!a, !b) = (undefined, undefined) in x + 1
Here, you create a tuple containing (a, b), and when the tuple is evaluated, both a and b are.
But because the tuple is never evaluated, neither a nor b are. This is basically the same as writing:
f x = let y = undefined `seq` 4 in x + 1
Since y is never evaluated, neither is undefined.
来源：https://stackoverflow.com/questions/6509399/tuple-bang-patterns