Returns an input stream for reading a resource in this module. The
name
parameter is a
'/'
-separated path name that identifies the resource. As with
Class.getResourceAsStream
, this method delegates to the module's class loader
findResource(String,String)
method, invoking it with the module name (or
null
when the module is unnamed) and the name of the resource. If the resource name has a leading slash then it is dropped before delegation.
A resource in a named module may be encapsulated so that it cannot be located by code in other modules. Whether a resource can be located or not is determined as follows:
- If the resource name ends with "
.class
" then it is not encapsulated.
- A package name is derived from the resource name. If the package name is a package in the module then the resource can only be located by the caller of this method when the package is open to at least the caller's module. If the resource is not in a package in the module then the resource is not encapsulated.
In the above, the package name for a resource is derived from the subsequence of characters that precedes the last '/'
in the name and then replacing each '/'
character in the subsequence with '.'
. A leading slash is ignored when deriving the package name. As an example, the package name derived for a resource named "a/b/c/foo.properties
" is "a.b.c
". A resource name with the name "META-INF/MANIFEST.MF
" is never encapsulated because "META-INF
" is not a legal package name.
This method returns null
if the resource is not in this module, the resource is encapsulated and cannot be located by the caller, or access to the resource is denied by the security manager.