Haskell infinite recursion

I've drawn a picture, which you might find helpful.
Note that zipWtih op (x:xs) (y:xs) = (op x y):zipWith xs ys, which is how zipWtih appears to "move" right along the list. It's reading elements and spitting out sums:

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Here's a more detailed step-by-step evaluation. (Although I'll paste copies of what's there, there's only one copy in memory.) I'll use .... for things I can't be bothered to write out.

fib = 0:1:zipWith (+) fib (tail fib)
    = 0:1:zipWith (+) (0:1: .... ) (tail (0:1: .... )
    = 0:1:(0+1:zipWith (+) (1:(0+1: .... )) ( 0+1:..... ))
    = 0:1:1:zipWith (+) (1: ....) (......)

notice that now we know that zipWith (+) fib (tail fib) = 1:......

    = 0:1:1:zipWith (+) (1:1: ....) (1:......)
    = 0:1:1:(1+1):zipWith (+) (1:(1+1): .....) ((1+1):....)
    = 0:1:1:2:zipWith (+) (1:2: .....) (2:....)

I'll go a little faster:

    = 0:1:1:2:(1+2):zipWith (+) (2: .....) (....)
    = 0:1:1:2:3     :zipWith (+) (2:3 .....) (3:....)
    = 0:1:1:2:3:(2+3):zipWith (+) (3:(2+3):.....) ((2+3):.....)
    = 0:1:1:2:3:5     :zipWith (+) (3:5:.....) (5:.....)
    = 0:1:1:2:3:5:8    :zipWith (+) (5:8:....) (8:......)
    = 0:1:1:2:3:5:8:13  :zipWith (+) (8:13:....) (13:......)
    = 0:1:1:2:3:5:8:13:21:zipWith (+) (13:21....) (21:......)

At each stage, the last two arguments to the zipWith function are like pointers to (one and two positions) further up the fib list than we are at present.

In a word: laziness. A list in Haskell is more like a generator: it will only compute values when they are demanded by something else.

For instance head [1 , 2+3] will not perform the addition, since it is not needed. Similarly, if we recursively let ones = 1 : ones, then head ones = head (1 : ones) = 1 does not need to evaluate all the tail.

You can try guessing what happens if we print a pair x, defined as follows:

x = (n, fst x + 1)

Above we use a (lazy) pair instead of a (lazy) list, but the reasoning is the same. Don't evaluate anything unless is it needed by something else.