java - example - ¿Cómo se diferencia Guava Splitter.onPattern(..). Split() de String.split(..)?

stream split string java 8

Question

Recientemente aproveché el poder de una expresión regular de mirar hacia adelante para dividir una cadena:

"abc8".split("(?=//d)|//W")

Si se imprime en la consola, esta expresión retorna:

[abc, 8]

Muy contento con este resultado, quería transferir esto a Guava para un mayor desarrollo, que se veía así:

Splitter.onPattern("(?=//d)|//W").split("abc8")

Para mi sorpresa, la salida cambió a:

[abc]

¿Por qué?

Answer 1

¡Has encontrado un error!

System.out.println(s.split("abc82")); // [abc, 8]
System.out.println(s.split("abc8")); // [abc]

Este es el método que utiliza Splitter para dividir String s ( Splitter.SplittingIterator::computeNext ):

@Override
protected String computeNext() {
/*
* The returned string will be from the end of the last match to the
* beginning of the next one. nextStart is the start position of the
* returned substring, while offset is the place to start looking for a
* separator.
*/
int nextStart = offset;
while (offset != -1) {
int start = nextStart;
int end;
int separatorPosition = separatorStart(offset);
if (separatorPosition == -1) {
end = toSplit.length();
offset = -1;
} else {
end = separatorPosition;
offset = separatorEnd(separatorPosition);
}
if (offset == nextStart) {
/*
* This occurs when some pattern has an empty match, even if it
* doesn''t match the empty string -- for example, if it requires
* lookahead or the like. The offset must be increased to look for
* separators beyond this point, without changing the start position
* of the next returned substring -- so nextStart stays the same.
*/
offset++;
if (offset >= toSplit.length()) {
offset = -1;
}
continue;
}
while (start < end && trimmer.matches(toSplit.charAt(start))) {
start++;
}
while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
end--;
}
if (omitEmptyStrings && start == end) {
// Don''t include the (unused) separator in next split string.
nextStart = offset;
continue;
}
if (limit == 1) {
// The limit has been reached, return the rest of the string as the
// final item. This is tested after empty string removal so that
// empty strings do not count towards the limit.
end = toSplit.length();
offset = -1;
// Since we may have changed the end, we need to trim it again.
while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
end--;
}
} else {
limit--;
}
return toSplit.subSequence(start, end).toString();
}
return endOfData();
}

El área de interés es:

if (offset == nextStart) {
/*
* This occurs when some pattern has an empty match, even if it
* doesn''t match the empty string -- for example, if it requires
* lookahead or the like. The offset must be increased to look for
* separators beyond this point, without changing the start position
* of the next returned substring -- so nextStart stays the same.
*/
offset++;
if (offset >= toSplit.length()) {
offset = -1;
}
continue;
}

Esta lógica funciona muy bien, a menos que la coincidencia vacía ocurra al final de una String . Si la coincidencia vacía ocurre al final de una String , terminará omitiendo ese carácter. Cómo debe ser esta parte (aviso >= - > ):

if (offset == nextStart) {
/*
* This occurs when some pattern has an empty match, even if it
* doesn''t match the empty string -- for example, if it requires
* lookahead or the like. The offset must be increased to look for
* separators beyond this point, without changing the start position
* of the next returned substring -- so nextStart stays the same.
*/
offset++;
if (offset > toSplit.length()) {
offset = -1;
}
continue;
}

Answer 2

El Splitter guayaba parece tener un error cuando un patrón coincide con una cadena vacía. Si intenta crear un Matcher e imprime lo que coincide:

Pattern pattern = Pattern.compile("(?=//d)|//W");
Matcher matcher = pattern.matcher("abc8");
while (matcher.find()) {
System.out.println(matcher.start() + "," + matcher.end());
}

Obtienes la salida 3,3 que hace que parezca que coincidiría con el 8 . Por lo tanto, simplemente se divide allí dando como resultado solo abc .

Puede usar, por ejemplo, Pattern#split(String) que parece dar el resultado correcto:

Pattern.compile("(?=//d)|//W").split("abc8")