I'm wondering what the objections are to using what I'll call the 'String constructor method' to convert an InputStream into a String.
Edit: added emphasis. In particular, I'm wondering why we have to mess with Streams and Buffers and Scanners and whatnot when this method seems to work fine.
private String readStream(InputStream in) {
byte[] buffer = new byte[1024];
try {
return new String(buffer, 0, in.read(buffer));
} catch (IOException e) {
Log.d(DEBUG_TAG, "Error reading input stream!");
return "";
}
}
I've seen this other helpful post and tried the methods I could:
Method 1, Apache commons, is a no-go, since I can't use and don't want libraries right now.
Method 2, The Scanner one, looks promising, but then you'd have to be able to set delimiters in the stream, which isn't always possible, right? E.g. right now I'm using an InputStream from a web API.
Method 3, the InputStreamReader in the slurp function, didn't work either - it gives me a bunch of numbers, where I'm sending a string with all types of characters, so I may be messing something up in my encoding.
But after many Google searches, I finally found the String constructor method, which is the only one that works for me.
From comments on the thread I linked, I know there are issues with encoding in the method I'm using. I've been coding for a while now and know what encodings are and why they're around. But I still lack any knowledge about what kinds of encodings are used where, and how to detect and handle them. Any resources/help on that topic would also be very appreciated!
Here is one method using only standard libraries:
use a ByteArrayOutputStream and copy all the bytes you receive in it;
wrap this ByteArrayOutputStream's bytes into a ByteBuffer;
use a CharsetDecoder to decode the ByteBuffer into a CharBuffer;
.toString() the CharBuffer after rewinding it.
Code (note: doesn't handle closing the input):
// Step 1: read all the bytes
final ByteArrayOutputStream out = new ByteArrayOutputStream();
final byte[] buffer = new byte[8196];
int count;
while ((count = in.read(buffer)) != -1)
out.write(buf, 0, count);
// Step 2: wrap the array
final ByteBuffer byteBuffer = ByteBuffer.wrap(out.toByteArray());
// Step 3: decode
final CharsetDecoder decoder = StandardCharsets.UTF_8.newDecoder()
.onUnmappableCharacter(CodingErrorAction.REPORT)
.onMalformedInput(CodingErrorAction.REPORT);
final CharBuffer charBuffer = decoder.decode(byteBuffer);
charBuffer.flip();
return charBuffer.toString();
Related
Given a Reader, a Charset, and a Path, how do I correctly and efficiently write the reader's content into a file?
The total size of the reader's content is not known in advance.
This is my current solution:
CharBuffer charBuffer = CharBuffer.allocate(1024);
try (FileChannel fc = (FileChannel) Files.newByteChannel(path, StandardOpenOption.WRITE, StandardOpenOption.CREATE_NEW)) {
while (true) {
int size = reader.read(charBuffer);
if (size < 0) break;
charBuffer.flip();
ByteBuffer bytes = charset.encode(charBuffer);
fc.write(bytes);
charBuffer.flip();
}
}
It works but it allocates a new ByteBuffer in every loop. I could try to reuse the byte buffer, but I would actually prefer a solution that uses only one buffer in total.
Using ByteBuffer#toCharBuffer is not an option because it does not consider the charset.
I also don't like the type cast in the try-statement, is there a cleaner solution?
The simplest way to transfer reader to a path is to use the built in methods of Files:
try(var out = Files.newBufferedWriter(path, charset, StandardOpenOption.WRITE, StandardOpenOption.CREATE_NEW)) {
reader.transferTo(out);
}
This does not need the CharBuffer and simplifies the logic of the code you need to write for this often needed task.
//Reading a image file from #drawable res folder and writing to a file on external sd card
//below one works no doubt but I want to imrpove it:
OutputStream os = new FileOutputStream(file); //File file.........
InputStream is =getResources().openRawResource(R.drawable.an_image);
byte[] b = new byte[is.available()];
is.read(b);
os.write(b);
is.close();
os.close();
In above code I am using basic io classes to read and write. My question is what can I do in order to able to use wrapper classes like say DataInputStream/ BufferedReaderd or PrintStream / BufferedWriter /PrintWriter.
As openRawResources(int id ) returns InputStream ;
to read a file from res I either need to typecast like this:
DataInputStream is = (DataInputStream) getResources().openRawResource(R.drawble.an_image));
or I can link the stream directly like this:
DataInputStream is = new DataInputStream(getResources().openRawResource(R.drawable.greenball));
and then I may do this to write it to a file on sd card:
PrintStream ps =new PrintStream (new FileOutputStream(file));
while(s=is.readLine()!=null){
ps.print(s);
}
So is that correct approach ? which one is better? Is there a better way?better practice..convention?
Thanks!!!
If openRawResource() is documented to return an InputStream then you cannot rely on that result to be any more specific kind of InputStream, and in particular, you cannot rely on it to be a DataInputStream. Casting does not change that; it just gives you the chance to experience interesting and exciting exceptions. If you want a DataInputStream wrapping the the result of openRawResource() then you must obtain it via the DataInputStream constructor. Similarly for any other wrapper stream.
HOWEVER, do note that DataInputStream likely is not the class you want. It is appropriate for reading back data that were originally written via a DataOutputStream, but it is inappropriate (or at least offers no advantages over any other InputStream) for reading general data.
Furthermore, your use of InputStream.available() is incorrect. That method returns the number of bytes that can currently be read from the stream without blocking, which has only a weak relationship with the total number of bytes that could be read from the stream before it is exhausted (if indeed it ever is).
Moreover, your code is also on shaky ground where it assumes that InputStream.read(byte[]) will read enough bytes to fill the array. It probably will, since that many bytes were reported available, but that's not guaranteed. To copy from one stream to another, you should instead use code along these lines:
private final static int BUFFER_SIZE = 2048;
void copyStream(InputStream in, OutputStream out) throws IOException {
byte[] buffer = new byte[BUFFER_SIZE];
int nread;
while ( (nread = in.read(buffer) != 0 ) do {
out.write(buffer, 0, nread);
}
}
I want to read a file into a byte array. So, I am reading it using:
int len1 = (int)(new File(filename).length());
FileInputStream fis1 = new FileInputStream(filename);
byte buf1[] = new byte[len1];
fis1.read(buf1);
However, it is realy very slow. Can anyone inform me a very fast approach (possibly best one) to read a file into byte array. I can use java library also if needed.
Edit: Is there any benchmark which one is faster (including library approach).
It is not very slow, at least there is not way to make it faster. BUT it is wrong. If file is big enough the method read() will not return all bytes from fist call. This method returns number of bytes it managed to read as return value.
The right way is to call this method in loop:
public static void copy(InputStream input,
OutputStream output,
int bufferSize)
throws IOException {
byte[] buf = new byte[bufferSize];
int bytesRead = input.read(buf);
while (bytesRead != -1) {
output.write(buf, 0, bytesRead);
bytesRead = input.read(buf);
}
output.flush();
}
call this as following:
ByteArrayOutputStream baos = new ByteArrayOutputStream();
copy(new FileInputStream(myfile), baos);
byte[] bytes = baos.toByteArray();
Something like this is implemented in a lot of packages, e.g. FileUtils.readFileToByteArray() mentioned by #Andrey Borisov (+1)
EDIT
I think that reason for slowness in your case is the fact that you create so huge array. Are you sure you really need it? Try to re-think your design. I believe that you do not have to read this file into array and can process data incrementally.
apache commons-io FileUtils.readFileToByteArray
I know how to get the inputstream for a given classpath resource, read from the inputstream until i reach the end, but it looks like a very common problem, and i wonder if there an API that I don't know, or a library that would make things as simple as
byte[] data = ResourceUtils.getResourceAsBytes("/assets/myAsset.bin")
or
byte[] data = StreamUtils.readStreamToEnd(myInputStream)
for example!
Java 9 native implementation:
byte[] data = this.getClass().getClassLoader().getResourceAsStream("/assets/myAsset.bin").readAllBytes();
Have a look at Google guava ByteStreams.toByteArray(INPUTSTREAM), this is might be what you want.
Although i agree with Andrew Thompson, here is a native implementation that works since Java 7 and uses the NIO-API:
byte[] data = Files.readAllBytes(Paths.get(this.getClass().getClassLoader().getResource("/assets/myAsset.bin").toURI()));
Take a look at Apache IOUtils - it has a bunch of methods to work with streams
I usually use the following two approaches to convert Resource into byte[] array.
1 - approach
What you need is to first call getInputStream() on Resource object, and then pass that to convertStreamToByteArray method like below....
InputStream stream = resource.getInputStream();
long size = resource.getFile().lenght();
byte[] byteArr = convertStreamToByteArray(stream, size);
public byte[] convertStreamToByteArray(InputStream stream, long size) throws IOException {
// check to ensure that file size is not larger than Integer.MAX_VALUE.
if (size > Integer.MAX_VALUE) {
return new byte[0];
}
byte[] buffer = new byte[(int)size];
ByteArrayOutputStream os = new ByteArrayOutputStream();
int line = 0;
// read bytes from stream, and store them in buffer
while ((line = stream.read(buffer)) != -1) {
// Writes bytes from byte array (buffer) into output stream.
os.write(buffer, 0, line);
}
stream.close();
os.flush();
os.close();
return os.toByteArray();
}
2 - approach
As Konstantin V. Salikhov suggested, you could use org.apache.commons.io.IOUtils and call its IOUtils.toByteArray(stream) static method and pass to it InputStream object like this...
byte[] byteArr = IOUtils.toByteArray(stream);
Note - Just thought I'll mention this that under the hood toByteArray(...) checks to ensure that file size is not larger than Integer.MAX_VALUE, so you don't have to check for this.
Commonly Java methods will accept an InputStream. In that majority of cases, I would recommend passing the stream directly to the method of interest.
Many of those same methods will also accept an URL (e.g. obtained from getResource(String)). That can sometimes be better, since a variety of the methods will require a repositionable InputStream and there are times that the stream returned from getResourceAsStream(String) will not be repositionable.
InputStream data = realResponse.getEntity().getContent();
byte[] preview = new byte[100];
data.read(preview, 0, 100);
// Now I want to refer to the InputStream later on, but I want it from the beginning of the stream, not 100 bytes in. I tried mark() it at 100, and then reset() after I read the first 100 bytes, but that doesn't work either.
Any ideas? Probably a stupid mistake..just not seeing it.
When you use mark() of the java.io.InputStream object you should check with the markSupported() method if your InputStream actually support using mark. According to the API the InputStream class doesn't, but the java.io.BufferedInputStream class does. Maybe you should embed your stream inside a BufferedInputStream object like:
InputStream data = new BufferedInputStream(realResponse.getEntity().getContent());
// data.markSupported() should return "true" now
data.mark(some_size);
// work with "data" now
...
data.reset();
If the InputStream supports mark (you can check with the markSupported() method), then the following should work:
InputStream data = realResponse.getEntity().getContent();
byte[] preview = new byte[100];
data.mark(100);
data.read(preview, 0, 100);
data.reset();
However, be aware that data.read(preview, 0, 100) is not guaranteed to read 100 bytes in one go, it may read less.