Arrays.copyOfRange( byte[], long, long ) ? in Java

Question

How can I 'cut' byte array using Arrays.copyOfRange with long parameters?

EDIT: I need to cut file ( not larger than 2GB ) from byte index to the end of file ( byte[].length )

int size = f.length()

throws: incompatible types: possible lossy conversion from long to int (even if the file is 600kB)

File f = new File("path-to-file");
byte[] bytes = Files.readAllBytes( f.toPath() );

long indexToCutFrom = a_long_number; 
long indexToCutTo = f.length() - 1;

// method with these parameters does not exists
byte[] cuttedBytes = Arrays.copyOfRange( bytes, indexToCutFrom, indexToCutTo );

How do you have an array with more than `Integer.MAX_VALUE` elements? — bradimus, Jan 27 '17 at 15:00
Arrays are limited to a length of `Integer.MAX_VALUE` so 2^32 -1. No need of `long` here. So in short, you can't because you don't need to — AxelH, Jan 27 '17 at 15:01
@Eugene, can you explain more about this? I might have right an answer to quickly ... — AxelH, Jan 27 '17 at 15:04
Are you sure `int` would not be enough? What's the reason you want to keep a byte array larger than 2 GB in memory? — Thomas Fritsch, Jan 27 '17 at 15:05
So, is your question how to cast/convert a `long` to an `int`? — bradimus, Jan 27 '17 at 15:10
See http://stackoverflow.com/questions/1590831/safely-casting-long-to-int-in-java — bradimus, Jan 27 '17 at 15:16

score 0 · Answer 1 · answered Jan 27 '17 at 15:03

0

Arrays are limited to a length of Integer.MAX_VALUE (= 2 ^ 32 -1). No need of long.

In short, you can't because you don't need to.

You can see it simply because Arrays.length() is an integer.

answered Jan 27 '17 at 15:03

AxelH

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Arrays.copyOfRange( byte[], long, long ) ? in Java

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