Get file name from FileOutputStream

Looks like the answer is no: http://download.oracle.com/javase/1.4.2/docs/api/java/io/FileOutputStream.html http://docs.oracle.com/javase/7/docs/api/index.html?java/io/FileOutputStream.html There are no public methods that return the File or String used in construction of the stream. EDIT: The same holds for FileInputStream.

Java Read Large Text File With 70million line of text

1) I am sure there is no difference speedwise, both use FileInputStream internally and buffering 2) You can take measurements and see for yourself 3) Though there’s no performance benefits I like the 1.7 approach try (BufferedReader br = Files.newBufferedReader(Paths.get(“test.txt”), StandardCharsets.UTF_8)) { for (String line = null; (line = br.readLine()) != null;) { // } … Read more

Buffered files (for faster disk access)

Windows file caching is very effective, especially if you are using Vista or later. TFileStream is a loose wrapper around the Windows ReadFile() and WriteFile() API functions and for many use cases the only thing faster is a memory mapped file. However, there is one common scenario where TFileStream becomes a performance bottleneck. That is … Read more

What’s the best way to open and read a file in Perl?

There are no universal standards, but there are reasons to prefer one or another. My preferred form is this: open( my $input_fh, “<“, $input_file ) || die “Can’t open $input_file: $!”; The reasons are: You report errors immediately. (Replace “die” with “warn” if that’s what you want.) Your filehandle is now reference-counted, so once you’re … Read more

How to use readline() method in Java?

I advise you to go with Scanner instead of DataInputStream. Scanner is specifically designed for this purpose and introduced in Java 5. See the following links to know how to use Scanner. Java Documentation Java Tutorial Example Scanner s = new Scanner(System.in); System.out.println(s.nextInt()); System.out.println(s.nextInt()); System.out.println(s.next()); System.out.println(s.next());

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