Java IO.pdf - Nguyen Dang Binh

Java I/O
try {
byte[] b = new byte[100];
int offset = 0;
while (offset < b.length) {
int bytesRead = System.in.read(b, offset, b.length - offset);
if (bytesRead == -1) break; // end of stream
offset += bytesRead;
}
}
catch (IOException e) {System.err.println("Couldn't read from
System.in!");}
3.4 Counting the Available Bytes
It's sometimes convenient to know how many bytes are available to be read before you
attempt to read them. The InputStream class's available() method tells you how many
bytes you can read without blocking. It returns if there's no data available to be read.
public int available() throws IOException
For example:
try {
byte[] b = new byte[100];
int offset = 0;
while (offset < b.length) {
int a = System.in.available();
int bytesRead = System.in.read(b, offset, a);
if (bytesRead == -1) break; // end of stream
offset += bytesRead;
}
catch (IOException e) {System.err.println("Couldn't read from
System.in!");}
There's a potential bug in this code. There may be more bytes available than there's space in
the array to hold them. One common idiom is to size the array according to the number
available() returns, like this:
try {
byte[] b = new byte[System.in.available()];
System.in.read(b);
}
catch (IOException e) {System.err.println("Couldn't read from
System.in!");}
This works well if you're only going to perform a single read. For multiple reads, however,
the overhead of creating multiple arrays is excessive. You should probably reuse the array and
only create a new array if more bytes are available than will fit in the array.
The available() method in java.io.InputStream always returns 0. Subclasses are
supposed to override it, but I've seen a few that don't. You may be able to read more bytes
from the underlying stream without blocking than available() suggests; you just can't
guarantee that you can. If this is a concern, you can place input in a separate thread so that
blocked input doesn't block the rest of the program.
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