How to create a new java.io.File in memory?

You could use an in-memory filesystem, such as Jimfs

Here's a usage example from their readme:

FileSystem fs = Jimfs.newFileSystem(Configuration.unix());
Path foo = fs.getPath("/foo");
Files.createDirectory(foo);

Path hello = foo.resolve("hello.txt"); // /foo/hello.txt
Files.write(hello, ImmutableList.of("hello world"), StandardCharsets.UTF_8);

How can I create new File (from java.io) in memory , not in the hard disk?

Maybe you are confusing File and Stream:

• A File is an abstract representation of file and directory pathnames. Using a File object, you can access the file metadata in a file system, and perform some operations on files on this filesystem, like delete or create the file. But the File class does not provide methods to read and write the file contents.
• To read and write from a file, you are using a Stream object, like FileInputStream or FileOutputStream. These streams can be created from a File object and then be used to read from and write to the file.

You can create a stream based on a byte buffer which resides in memory, by using a ByteArrayInputStream and a ByteArrayOutputStream to read from and write to a byte buffer in a similar way you read and write from a file. The byte array contains the "File's" content. You do not need a File object then.

Both the File... and the ByteArray... streams inherit from java.io.OutputStream and java.io.InputStream, respectively, so that you can use the common superclass to hide whether you are reading from a file or from a byte array.

I think temporary file can be another solution for that.

File tempFile = File.createTempFile(prefix, suffix, null);
FileOutputStream fos = new FileOutputStream(tempFile);
fos.write(byteArray);

There is a an answer about that here.

It is not possible to create a java.io.File that holds its content in (Java heap) memory *.

Instead, normally you would use a stream. To write to a stream, in memory, use:

OutputStream out = new ByteArrayOutputStream();
out.write(...);

But unfortunately, a stream can't be used as input for java.util.jar.JarFile, which as you mention can only use a File or a String containing the path to a valid JAR file. I believe using a temporary file like you currently do is the only option, unless you want to use a different API.

If you are okay using a different API, there is conveniently a class in the same package, named JarInputStream you can use. Simply wrap your archiveContent array in a ByteArrayInputStream, to read the contents of the JAR and extract the manifest:

try (JarInputStream stream = new JarInputStream(new ByteArrayInputStream(archiveContent))) {
     Manifest manifest = stream.getManifest();
}

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*) It's obviously possible to create a full file-system that resides in memory, like a RAM-disk, but that would still be "on disk" (and not in Java heap memory) as far as the Java process is concerned.