Memory map directly to string

Question

I have a file I'm mapping into memory via 'FileChannel.map()'. However it seems a bit odd when reading a string to do the following:

1) read a int for the string length
2) allocate a byte[length] object
3) use .get to read length bytes
4) convert the byte[] to a string

Now I know from my C++ background that memory mapped files are given to the user as pointers to memory. So is there a good way to skip using a byte array and just have the string conversion go right off the mapped memory?

That's to be determined at this point. Currently it's ASCII. — Timothy Baldridge
– Timothy Baldridge, Commented Apr 29, 2011 at 19:54

Pih · Accepted Answer · 2011-04-29 20:00:38Z

3

I suggest:

MappedByteBuffer mapped = fileChannel.map(mode, position, size);
String s = new String(mapped.array());

It is also possible to use the mapped.asCharBuffer() and get the chars by this way.

answered Apr 29, 2011 at 20:00

Pih

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3 Comments

Steve Blackwell Over a year ago

To prevent the unnecessary copy to a buffer, this is the way to do it. Of course, making a string still involves a copy, as rlibby mentioned.

rlibby Over a year ago

...so that he can get an UnsupportedOperationException? MappedByteBuffer is not backed by an array, MappedByteBuffer.hasArray() returns false, and MappedByteBuffer.array() throws the above exception.

Pih Over a year ago

It could be true, so, he will need to use asCharBuffer() and read each char to another buffer and so on construct the string.

ccleve · Accepted Answer · 2011-04-29 19:50:40Z

3

Ultimately, no. But there is a way to get a view of the data as characters. Look at ByteBuffer.asCharBuffer().

Internally, the asCharBuffer() method does the same thing you're proposing, but on a char-by-char basis.

answered Apr 29, 2011 at 19:50

ccleve

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1 Comment

BlueRaja - Danny Pflughoeft Over a year ago

+1 that is the best he could possibly hope for, if the file is ASCII, because Java strings are Unicode..

rlibby · Accepted Answer · 2011-04-29 19:50:56Z

1

There's no getting around String wanting a private copy of the data. Strings are immutable and if it used a shared array, you could break that. It's unfortunate that there's no String(CharSequence) or String(ByteBuffer) constructor.

answered Apr 29, 2011 at 19:50

rlibby

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