Get Linux file descriptor of FileChannel

Is there a way to get the Linux file descriptor for an opened FileChannel?

I need it to call mount.fuse -o fd=... (for implementing FUSE).

As a hacky workaround, I'm doing:

var pid = ProcessHandle.current().pid();
var fd = Files.list(Path.of("/proc/"+pid+"/fd")).count();
var fc = FileChannel.open(path);
System.out.println("file descriptor: " + fd);

Note that two file descriptors appear. One for path and another socket. I'm using the first one. What is the socket for?

You can use reflection to get the file descriptor from a RandomAccessFile:

long pid = ProcessHandle.current().pid();
long guessedFd = Files.list(Path.of("/proc/"+pid+"/fd")).count();

var file = new java.io.RandomAccessFile(FUSE_DEVICE_PATH.toFile(), "rw");
FileChannel fc = file.getChannel(); // Use FileChannel for fast NIO
var javaFd = file.getFD();
try {
    Field f = FileDescriptor.class.getDeclaredField("fd");
    f.setAccessible(true);
    var trueFd = (int) f.get(javaFd);
    return trueFd;
}
catch (InaccessibleObjectException | NoSuchFieldException | IllegalAccessException e) {
    return guessedFd;
}

On Java 11+ you will the jvm option --add-opens=java.base/java.io=ALL-UNNAMED at startup. Tested on Java 18.

In Java 17 you can use SharedSecrets instead of reflection:

FileDescriptor javaFd = ...
try {
    int trueFd = jdk.internal.access.SharedSecrets.getJavaIOFileDescriptorAccess().get(javaFd);
    return trueFd;
}
catch (IllegalAccessError e) {
    return guessedFd;
}

You will need to add --add-exports=java.base/jdk.internal.access=ALL-UNNAMED when compiling and running. See for Maven.