(* Substring -- SML Basis Library *) type substring val substring : string * int * int -> substringval extract   : string * int * int option -> substringval all       : string -> substringval string    : substring -> stringval base      : substring -> (string * int * int) val isEmpty   : substring -> boolval getc      : substring -> (char * substring) optionval first     : substring -> char optionval triml     : int -> substring -> substringval trimr     : int -> substring -> substringval sub       : substring * int -> charval size      : substring -> intval slice     : substring * int * int option -> substringval concat    : substring list -> stringval explode   : substring -> char listval isPrefix  : string -> substring -> boolval compare   : substring * substring -> orderval collate   : (char * char -> order) -> substring * substring -> order val dropl     : (char -> bool) -> substring -> substringval dropr     : (char -> bool) -> substring -> substringval takel     : (char -> bool) -> substring -> substringval taker     : (char -> bool) -> substring -> substringval splitl    : (char -> bool) -> substring -> substring * substringval splitr    : (char -> bool) -> substring -> substring * substringval splitAt   : substring * int -> substring * substring val position  : string -> substring -> substring * substring exception Spanval span      : substring * substring -> substring val translate : (char -> string) -> substring -> string val tokens    : (char -> bool) -> substring -> substring listval fields    : (char -> bool) -> substring -> substring list val foldl     : (char * 'a -> 'a) -> 'a -> substring -> 'aval foldr     : (char * 'a -> 'a) -> 'a -> substring -> 'aval app       : (char -> unit) -> substring -> unit (*    [substring] is the type of substrings of a basestring, an efficient    representation of a piece of a string.   A substring (s,i,n) is valid if 0 <= i <= i+n <= size s,                   or equivalently, 0 <= i and 0 <= n and i+n <= size s.     A valid substring (s, i, n) represents the string s[i...i+n-1].     Invariant in the implementation: Any value of type substring is valid.    [substring(s, i, n)] creates the substring (s, i, n), consisting of   the substring of s with length n starting at i.  Raises Subscript   if i<0 or n<0 or i+n > size s.  Equivalent to extract(s, i, SOME n).    [extract(s, i, NONE)] creates the substring (s, i, size s-i)   consisting of the tail of s starting at i.     Raises Subscript if i<0 or i > size s.    [extract(s, i, SOME n)] creates the substring (s, i, n),   consisting of the substring of s with length n starting at i.   Raises Subscript if i<0 or n<0 or i+n > size s.    [all s] is the substring (s, 0, size s).    [string sus] is the string s[i..i+n-1] represented by sus = (s, i, n).    [base sus] is the concrete triple (s, i, n), where sus = (s, i, n).    [isEmpty (s, i, n)] true if the substring is empty (that is, n = 0).    [getc sus] returns SOME(c, rst) where c is the first character and   rst the remainder of sus, if sus is non-empty; otherwise returns   NONE.  Note that         #1 o valOf o scanFn Substring.getc   is equivalent to, but more efficient than,         valOf o StringCvt.scanString scanFn o Substring.string    [first sus] returns SOME c where c is the first character in sus,   if sus is non-empty; otherwise returns NONE.    [triml k sus] returns sus less its leftmost k characters; or the   empty string at the end of sus if it has less than k characters.   Raises Subscript if k < 0, even in the partial application triml(k).     [trimr k sus] returns sus less its rightmost k characters; or the   empty string at the beginning of sus if it has less than k characters.     Raises Subscript if k < 0, even in the partial application triml(k).    [sub (sus, k)] returns the k'th character of the substring; that is,   s(i+k) where sus = (s, i, n).  Raises Subscript if k<0 or k>=n.    [size (s, i, n)] returns the size of the substring, that is, n.    [slice (sus, i', NONE)] returns the substring (s, i+i', n-i'), where   sus = (s, i, n).  Raises Subscript if i' < 0 or i' > n.    [slice (sus, i', SOME n')] returns the substring (s, i+i', n'), where   sus = (s, i, n).  Raises Subscript if i' < 0 or n' < 0 or i'+n' >= n.    [concat suss] returns a string consisting of the concatenation of   the substrings.  Equivalent to String.concat (List.map string suss).    [explode sus] returns the list of characters of sus, that is,        [s(i), s(i+1), ..., s(i+n-1)]   where sus = (s, i, n).  Equivalent to String.explode(string ss).    [isPrefix s1 s2] is true if s1 is a prefix of s2. That is, if there    exists a string t such that string s1 ^ t = string s2.    [compare (sus1, sus2)] performs lexicographic comparison, using the   standard ordering Char.compare on the characters.  Returns LESS,   EQUAL, or GREATER, according as sus1 is less than, equal to, or   greater than sus2.  Equivalent to, but more efficient than,       String.compare(string sus1, string sus2).    [collate cmp (sus1, sus2)] performs lexicographic comparison, using the    given ordering cmp on characters.  Equivalent to, but more efficient    than, String.collate cmp (string sus1, string sus2).    [dropl p sus] drops the longest prefix (left substring) of sus all   of whose characters satisfy predicate p.  If all characters do, it   returns the empty substring (s, i+n, 0) where sus = (s, i, n).    [dropr p sus] drops the longest suffix (right substring) of sus all   of whose characters satisfy predicate p.  If all characters do, it   returns the empty substring (s, i, 0) where sus = (s, i, n).    [takel p sus] returns the longest prefix (left substring) of sus   all of whose characters satisfy predicate p.  That is, if the   left-most character does not satisfy p, returns the empty (s, i, 0)   where sus = (s, i, n).    [taker p sus] returns the longest suffix (right substring) of sus   all of whose characters satisfy predicate p.  That is, if the   right-most character satisfies p, returns the empty (s, i+n, 0)   where sus = (s, i, n).    Let p be a predicate and xxxxfyyyyfzzzz a string where all   characters in xxxx and zzzz satisfy p, and f a is character   not satisfying p.  Then                 sus = xxxxfyyyyfzzzz         sus = xxxxzzzz        ------------------------------------------------------        dropl p sus =     fyyyyfzzzz                       dropr p sus = xxxxfyyyyf               takel p sus = xxxx                         xxxxzzzz        taker p sus =           zzzz               xxxxzzzz    It also holds that         concat[takel p sus, dropl p sus] = string sus        concat[dropr p sus, taker p sus] = string sus     [splitl p sus] splits sus into a pair (sus1, sus2) of substrings   where sus1 is the longest prefix (left substring) all of whose   characters satisfy p, and sus2 is the rest.  That is, sus2 begins   with the leftmost character not satisfying p.  Disregarding   sideeffects, we have:         splitl p sus = (takel p sus, dropl p sus).    [splitr p sus] splits sus into a pair (sus1, sus2) of substrings   where sus2 is the longest suffix (right substring) all of whose   characters satisfy p, and sus1 is the rest.  That is, sus1 ends   with the rightmost character not satisfying p.  Disregarding   sideeffects, we have:        splitr p sus = (dropr p sus, taker p sus)    [splitAt (sus, k)] returns the pair (sus1, sus2) of substrings,   where sus1 contains the first k characters of sus, and sus2   contains the rest.  Raises Subscript if k < 0 or k > size sus.    [position s (s',i,n)] splits the substring into a pair (pref, suff)   of substrings, where suff is the longest suffix of (s', i, n) which   has s as a prefix.  More precisely, let m = size s.  If there is a   least index k in i..i+n-m for which s = s'[k..k+m-1],    then the result is       pref = (s', i, k-i) and suff = (s', k, n-(k-i));    otherwise the result is  pref = (s', i, n)   and suff = (s', i+n, 0).    [span (sus1, sus2)] returns a substring spanning from the start of   sus1 to the end of sus2, provided this is well-defined: sus1 and   sus2 must have the same underlying string, and the start of sus1   must not be to the right of the end of sus2; otherwise raises Span.    More precisely, if base(sus1) = (s,i,n) and base(sus2) = (s',i',n')    and s = s' and i <= i'+n', then base(join(sus1, sus2)) = (s, i, i'+n'-i).   This may be used to compute `span', `union', and `intersection'.    [translate f sus] applies f to every character of sus, from left to   right, and returns the concatenation of the results.  Raises Size   if the sum of their sizes is greater than String.maxSize.   Equivalent to String.concat(List.map f (explode sus)).    [tokens p sus] returns the list of tokens in sus, from left to right,    where a token is a non-empty maximal substring of sus not containing    any delimiter, and a delimiter is a character satisfying p.    [fields p sus] returns the list of fields in sus, from left to right,    where a field is a (possibly empty) maximal substring of sus not    containing any delimiter, and a delimiter is a character satisfying p.    Two tokens may be separated by more than one delimiter, whereas two   fields are separated by exactly one delimiter.  If the only delimiter    is the character #"|", then        "abc||def" contains two tokens:   "abc" and "def"        "abc||def" contains three fields: "abc" and "" and "def"    [foldl f e sus] folds f over sus from left to right.  That is,    evaluates f(s[i+n-1], f( ... f(s[i+1], f(s[i] % e)) ...))    tail-recursively, where sus = (s, i, n).     Equivalent to List.foldl f e (explode sus).    [foldr f e sus] folds f over sus from right to left.  That is,    evaluates f(s[i], f(s[i+1], f(... f(s[i+n-1] % e) ...)))   tail-recursively, where sus = (s, i, n).   Equivalent to List.foldr f e (explode sus).    [app f sus] applies f to all characters of sus, from left to right.   Equivalent to List.app f (explode sus). *)