Like this: InputStream stream = new ByteArrayInputStream(exampleString.getBytes(StandardCharsets.UTF_8)); Note that this assumes that you want an InputStream that is a stream of bytes that represent your original string encoded as UTF-8. For versions of Java less than 7, replace StandardCharsets.UTF_8 with “UTF-8”.
The shortest solution I know: using(var memoryStream = new MemoryStream()) { sourceStream.CopyTo(memoryStream); return memoryStream.ToArray(); }
There seem to be many links and other such stuff, but no actual code using pipes. The advantage of using java.io.PipedInputStream and java.io.PipedOutputStream is that there is no additional consumption of memory. ByteArrayOutputStream.toByteArray() returns a copy of the original buffer, so that means that whatever you have in memory, you now have two copies of … Read more
Console.Read() reads only the next character from standard input, and Console.ReadLine() reads the next line of characters from the standard input stream. Standard input in case of Console Application is input from the user typed words in console UI of your application. Try to create it by Visual studio, and see by yourself.
Faced the same issue. I’ve cut up my 4MB file into 1 MB chunks, and on the first run I join the chunks into a data folder on the phone. As an added bonus, the APK is properly compressed. The chunk files are called 1.db, 2.db, etc. The code goes like this: File Path = … Read more
Try this : I assume your text file is on sd card //Find the directory for the SD Card using the API //*Don’t* hardcode “/sdcard” File sdcard = Environment.getExternalStorageDirectory(); //Get the text file File file = new File(sdcard,”file.txt”); //Read text from file StringBuilder text = new StringBuilder(); try { BufferedReader br = new BufferedReader(new FileReader(file)); … Read more
There are subtle differences as to how the fileName you are passing is interpreted. Basically, you have 2 different methods: ClassLoader.getResourceAsStream() and Class.getResourceAsStream(). These two methods will locate the resource differently. In Class.getResourceAsStream(path), the path is interpreted as a path local to the package of the class you are calling it from. For example calling, … Read more
Using inputStream.available() It is always acceptable for System.in.available() to return 0. I’ve found the opposite – it always returns the best value for the number of bytes available. Javadoc for InputStream.available(): Returns an estimate of the number of bytes that can be read (or skipped over) from this input stream without blocking by the next … Read more
I need to store data into files inside .jar file and read it again No you don’t. Instead store the ‘default’ file inside the Jar. If it is changed, store the altered file in another place. One common place is a sub-directory of user.home. When checking for the file, first check the existence of an … Read more
While Jon’s answer is correct, he is rewriting code that already exists in CopyTo. So for .Net 4 use Sandip’s solution, but for previous version of .Net use Jon’s answer. Sandip’s code would be improved by use of “using” as exceptions in CopyTo are, in many situations, quite likely and would leave the MemoryStream not … Read more