Haskell basic management instructions - a statement that always works?

Question

If I wanted to do the following:

function1 stuff
| condition1 = "yay" 
| condition2 = "hey" 
| condition3 = "nay" 

and I wanted to have a statement that was always executed unconditionally (for example, "because"), how could I do this? Lines are placeholders.

Thank!

Edit:

Before I give up on this, I want to try to rephrase it in terms of the problem.

I have a function that accepts two inputs (lists of lists).

If the input pattern is specified [[]] [[[]], [[]]] (or [[]] [[[x]], [[]]]), I want

• check if there is an even number of [[]] on each side, and if so remove them all
• check the odd amount on each side, then dial (delete all but one).
• several other similar things

() . , , , --. (), , =, . ... , , - haskell?

Answer 1

, Haskell. , , " A , B?". : " A, B?". ( OO), , . , , , ( , ).

,

number = number * 5
number = number * 2

" " " 5, " " 2".

( "" ), ( " * 5?" ).

, , , : , . , . , :

length []     = 0                 -- The empty list has zero items.
length (x:xs) = 1 + length xs     -- The non-empty list (x:xs) has one more item than xs.

, .

Answer 2

, , , otherwise:

foo x
    | x == 0 = "yay"
    | x == 1 = "hey"
    | x >= 2 = "nay"
    | otherwise = "negative"

- , " ", "", , , . , :

foo x
    | x == 1 = "one"
    | always = "any"

foo 1?

Answer 3

The sole purpose of a function is to return a value. If you, however, have some commonality that each condition shares, you can usewhere

function1 stuff
  | condition1 = "yay" ++ otherstuff
  | condition2 = "hey" ++ otherstuff
  | condition3 = "nay" ++ otherstuff
  where otherstuff = "!"