Is there an efficient way to create a file with a given size in Java?
In C it can be done with ftruncate (see that answer).
Most people would just write n dummy bytes into the file, but there must be a faster way. I'm thinking of ftruncate and also of Sparse files…
Create a new RandomAccessFile and call the setLength method, specifying the desired file length. The underlying JRE implementation should use the most efficient method available in your environment.
The following program
import java.io.*;
class Test {
public static void main(String args[]) throws Exception {
RandomAccessFile f = new RandomAccessFile("t", "rw");
f.setLength(1024 * 1024 * 1024);
}
}
on a Linux machine will allocate the space using the ftruncate(2)
6070 open("t", O_RDWR|O_CREAT, 0666) = 4
6070 fstat(4, {st_mode=S_IFREG|0644, st_size=0, ...}) = 0
6070 lseek(4, 0, SEEK_CUR) = 0
6070 ftruncate(4, 1073741824) = 0
while on a Solaris machine it will use the the F_FREESP64 function of the fcntl(2) system call.
/2: open64("t", O_RDWR|O_CREAT, 0666) = 14
/2: fstat64(14, 0xFE4FF810) = 0
/2: llseek(14, 0, SEEK_CUR) = 0
/2: fcntl(14, F_FREESP64, 0xFE4FF998) = 0
In both cases this will result in the creation of a sparse file.
Since Java 8, this method works on Linux and Windows :
final ByteBuffer buf = ByteBuffer.allocate(4).putInt(2);
buf.rewind();
final OpenOption[] options = { StandardOpenOption.WRITE, StandardOpenOption.CREATE_NEW , StandardOpenOption.SPARSE };
final Path hugeFile = Paths.get("hugefile.txt");
try (final SeekableByteChannel channel = Files.newByteChannel(hugeFile, options);) {
channel.position(HUGE_FILE_SIZE);
channel.write(buf);
}
You can open the file for writing, seek to offset (n-1), and write a single byte. The OS will automatically extend the file to the desired number of bytes.
