On 24 June 2011 10:31, divyanshu ranjan
My question is why it produces list in first place. 1:2:3 is not a list but 1:2:3:[] is. How comes haskell know that the things which we are adding will eventually added to the beginning of empty list given things are infinite so you can specify the end which is [] . Then why doing 1:3:4 is not acceptable ?
It's quite simple if you consider that it's just based on types of expressions and nothing more. Consider the type of the (:) function: (:) :: x -> [x] -> [x] It takes x and list of x. So we can write: () : [] And the expression () : [] has type [()] Therefore we can also take that and put it on the right-hand side again: () : (() : []) See how () and (() : []) fit into the x and [x] places in the (:) function? Continuing: () : (() : (() : [])) And so on. The (:) function is right-associative (meaning x : b : c : d means x : (b : (c : d))). So we don't need the parentheses, the following is equivalent to the previous: () : () : () : [] The a:b:c syntax isn't special in any way. But you should always implicitly read it as (a:(b:c)), remembering that everything on the left-hand side of a : is type x, and everything on right-hand side is [x]. repeat' x has type [x], so you can write x : repeat x E.g. consider manually evaluating repeat () on paper, we'd write out: repeat () () : repeat () () : () : repeat () () : () : () : repeat () And so on. It will go on forever if you keep evaluating the end expression. Haskell is lazy, so we don't have to evaluate the right-hand side if we don't want use it. But the type of the expressions, you should observe, are always correct. The reason 1:3:4 is not acceptable is because 4 is not of type [x], whereas 1:3:repeat 4 would be because repeat 4 is of type [x]. Ciao!