如何避免制作ByteBuffer的防御性副本?
2022-09-02 14:03:07
我有一个类,它采用ByteBuffer作为构造函数参数。有没有办法避免制作防御性副本,以确保缓冲区不会在该点之后被修改?
ByteBuffer.isReadOnly() 不保证原始所有者不会修改缓冲区。更糟糕的是,似乎没有办法对ByteBuffer进行子类化。有什么想法吗?
我有一个类,它采用ByteBuffer作为构造函数参数。有没有办法避免制作防御性副本,以确保缓冲区不会在该点之后被修改?
ByteBuffer.isReadOnly() 不保证原始所有者不会修改缓冲区。更糟糕的是,似乎没有办法对ByteBuffer进行子类化。有什么想法吗?
正如你所说,唯一真正的方法是,然后把它传递到构造函数中。除此之外,没有其他选择,尽管您可以将内容复制到新的,然后传递它。buf.asReadOnlyBuffer()
ByteBuffer
这是我现在能做的最好的事情:
/**
* Helper functions for java.nio.Buffer.
* <p/>
* @author Gili Tzabari
*/
public final class Buffers
{
/**
* Returns a ByteBuffer that is identical but distinct from the original buffer.
* <p/>
* @param original the buffer to copy
* @return an independent copy of original
* @throws NullPointerException if original is null
*/
public static ByteBuffer clone(ByteBuffer original)
{
Preconditions.checkNotNull(original, "original may not be null");
ByteBuffer result = ByteBuffer.allocate(original.capacity());
ByteBuffer source = original.duplicate();
source.rewind();
result.put(source);
try
{
source.reset();
result.position(source.position());
result.mark();
}
catch (InvalidMarkException unused)
{
// Mark is unset, ignore.
}
result.position(original.position());
result.limit(original.limit());
return result;
}
/**
* Returns an array representation of a buffer. The returned buffer may, or may not, be tied to
* the underlying buffer's contents (so it should not be modified).
* <p/>
* @param buffer the buffer
* @return the remaining bytes
*/
public static byte[] toArray(ByteBuffer buffer)
{
if (buffer.hasArray() && !buffer.isReadOnly() && buffer.position() == 0
&& buffer.remaining() == buffer.limit())
{
return buffer.array();
}
ByteBuffer copy = buffer.duplicate();
byte[] result = new byte[copy.remaining()];
copy.get(result);
return result;
}
/**
* Prevent construction.
*/
private Buffers()
{
}
}
我还向Oracle提交了功能请求:http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=7130631