A path pattern, matched against a whole target string: a tool name, a /-separated path, or a URL.
* matches any run of characters other than /, so it stays within one segment;
? matches one character other than /;
** matches any run of characters, / included; as a whole segment it also matches zero segments, so src/**/a.scala matches src/a.scala, and src/** matches src.
Every other character is literal: there are no escapes, character classes or braces, so every string is a valid pattern. Two globs are equal when their patterns are.
Besides matching one target, two globs compare as sets of targets: overlaps (some target matches both) and covers (every target the other matches, this one matches). A permission check uses them for a resource whose target is itself a pattern, an envelope of the targets a call may touch. Matching and both comparisons run on one automaton built from the pattern, so they cannot disagree.
It is reflexive: for any instance x of type Any, x.equals(x) should return true.
It is symmetric: for any instances x and y of type Any, x.equals(y) should return true if and only if y.equals(x) returns true.
It is transitive: for any instances x, y, and z of type Any if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.
If you override this method, you should verify that your implementation remains an equivalence relation. Additionally, when overriding this method it is usually necessary to override hashCode to ensure that objects which are "equal" (o1.equals(o2) returns true) hash to the same scala.Int. (o1.hashCode.equals(o2.hashCode)).
Value parameters
that
the object to compare against this object for equality.
Attributes
Returns
true if the receiver object is equivalent to the argument; false otherwise.
The default hashing algorithm is platform dependent.
Note that it is allowed for two objects to have identical hash codes (o1.hashCode.equals(o2.hashCode)) yet not be equal (o1.equals(o2) returns false). A degenerate implementation could always return 0. However, it is required that if two objects are equal (o1.equals(o2) returns true) that they have identical hash codes (o1.hashCode.equals(o2.hashCode)). Therefore, when overriding this method, be sure to verify that the behavior is consistent with the equals method.