Haskell无限循环，简单地重新放行动

Question

I have such code:

Prelude> let n = [1,2,3,4]
Prelude> n
[1,2,3,4]
Prelude> 0:n
[0,1,2,3,4]
Prelude> let n = 0:n

And when I type in Haskell interpreter after upper:

Prelude> n

I'm getting the infinite result:

[0,0,0,0,0,0,0,0,0

And where printing " 0, " is infinite.

Why do I get such result?
Is there some recursive stuff, and why/how does it work in interpreater level?
Could I catch stack overflow, which such stuff on GHCi or not?

Thanks,
Best Regards!

Answer 1

The new binding of n shadows the old one. You don't reassign to variables in Haskell.

Answer 2

What Josh is saying is that you definition of n expands as:

0:n.  -- note n still equals 0:n, just like you said
0:0:n. -- note n _still_ equals 0:n
0:0:0:n
...

Answer 3

Haskell's let is like letrec in other languages such as ML: bindings are allowed to be recursive. In other words n = 0:n defines n to be a list where the first element is 0 and the rest of the list is equal to n . That means that the second element of n is equal to the first element of n , which is 0 , etc.

Infinite lists are OK in Haskell due to laziness. If you only ever use the first 10 elements of the list, then the 11th element and beyond will never be evaluated.