O(1) Returns the first character of a Text, which must be non-empty. Subject to fusion.

O(1) Returns the last character of a Text, which must be non-empty. Subject to fusion.

O(1) Returns all characters after the head of a Text, which must be non-empty. Subject to fusion.

O(1) Returns all but the last character of a Text, which must be non-empty. Subject to fusion.

O(m+n) Replace every non-overlapping occurrence of needle in haystack with replacement.

This function behaves as though it was defined as follows:

replace needle replacement haystack =
  intercalate replacement (splitOn needle haystack)

As this suggests, each occurrence is replaced exactly once. So if needle occurs in replacement, that occurrence will not itself be replaced recursively:

>>> replace "oo" "foo" "oo"
"foo"

In cases where several instances of needle overlap, only the first one will be replaced:

>>> replace "ofo" "bar" "ofofo"
"barfo"

In (unlikely) bad cases, this function's time complexity degrades towards O(n*m).

O(n) A variant of foldl that has no starting value argument, and thus must be applied to a non-empty Text. Subject to fusion.

O(n) A strict version of foldl1. Subject to fusion.

O(n) A variant of foldr that has no starting value argument, and thus must be applied to a non-empty Text. Subject to fusion.

O(n) maximum returns the maximum value from a Text, which must be non-empty. Subject to fusion.

O(n) minimum returns the minimum value from a Text, which must be non-empty. Subject to fusion.

O(n+m) Find the first instance of needle (which must be non-null) in haystack. The first element of the returned tuple is the prefix of haystack before needle is matched. The second is the remainder of haystack, starting with the match.

Examples:

>>> breakOn "::" "a::b::c"
("a","::b::c")

>>> breakOn "/" "foobar"
("foobar","")

Laws:

append prefix match == haystack
  where (prefix, match) = breakOn needle haystack

If you need to break a string by a substring repeatedly (e.g. you want to break on every instance of a substring), use breakOnAll instead, as it has lower startup overhead.

In (unlikely) bad cases, this function's time complexity degrades towards O(n*m).

O(n+m) Similar to breakOn, but searches from the end of the string.

The first element of the returned tuple is the prefix of haystack up to and including the last match of needle. The second is the remainder of haystack, following the match.

>>> breakOnEnd "::" "a::b::c"
("a::b::","c")

O(m+n) Break a Text into pieces separated by the first Text argument (which cannot be empty), consuming the delimiter. An empty delimiter is invalid, and will cause an error to be raised.

Examples:

>>> splitOn "\r\n" "a\r\nb\r\nd\r\ne"
["a","b","d","e"]

>>> splitOn "aaa"  "aaaXaaaXaaaXaaa"
["","X","X","X",""]

>>> splitOn "x"    "x"
["",""]

and

intercalate s . splitOn s         == id
splitOn (singleton c)             == split (==c)

(Note: the string s to split on above cannot be empty.)

In (unlikely) bad cases, this function's time complexity degrades towards O(n*m).

O(n+m) Find all non-overlapping instances of needle in haystack. Each element of the returned list consists of a pair:

• The entire string prior to the kth match (i.e. the prefix)
• The kth match, followed by the remainder of the string

Examples:

>>> breakOnAll "::" ""
[]

>>> breakOnAll "/" "a/b/c/"
[("a","/b/c/"),("a/b","/c/"),("a/b/c","/")]

In (unlikely) bad cases, this function's time complexity degrades towards O(n*m).

The needle parameter may not be empty.

O(n+m) The count function returns the number of times the query string appears in the given Text. An empty query string is invalid, and will cause an error to be raised.

In (unlikely) bad cases, this function's time complexity degrades towards O(n*m).