Fastest way to read huge number of int from binary file

Question

I use Java 1.5 on an embedded Linux device and want to read a binary file with 2MB of int values. (now 4bytes Big Endian, but I can decide, the format)

Using D

Answer 1

You can use IntBuffer from nio package -> http://docs.oracle.com/javase/6/docs/api/java/nio/IntBuffer.html

int[] intArray = new int[ 5000000 ];

IntBuffer intBuffer = IntBuffer.wrap( intArray );

...

Fill in the buffer, by making calls to inChannel.read(intBuffer).

Once the buffer is full, your intArray will contain 500000 integers.

EDIT

After realizing that Channels only support ByteBuffer.

// asume I already know that there are now 500 000 int to read:
int numInts = 500000;
// here I want the result into
int[] result = new int[numInts];

// 4 bytes per int, direct buffer
ByteBuffer buf = ByteBuffer.allocateDirect( numInts * 4 );

// BIG_ENDIAN byte order
buf.order( ByteOrder.BIG_ENDIAN );

// Fill in the buffer
while ( buf.hasRemaining( ) )
{
   // Per EJP's suggestion check EOF condition
   if( inChannel.read( buf ) == -1 )
   {
       // Hit EOF
       throw new EOFException( );
   }
}

buf.flip( );

// Create IntBuffer view
IntBuffer intBuffer = buf.asIntBuffer( );

// result will now contain all ints read from file
intBuffer.get( result );