[Java] MemoryBuffer::readBytesAsInt64 with len=0 Causes '>>>' Shift Operation by Overly Large Constant Value Issue

[LouisLou2]

Search before asking

• [x] I had searched in the issues and found no similar issues.

Version

Fury Java 0.10.0

Component(s)

Java

Minimal reproduce step

1. Use the following method in your Java code:

   /** Read {@code len} bytes into a long using little-endian order. */
   public long readBytesAsInt64(int len) {
       int readerIdx = readerIndex;
       int remaining = size - readerIdx;
       if (remaining >= 8) {
           readerIndex = readerIdx + len;
           long v = UNSAFE.getLong(heapMemory, address + readerIdx);
           v = (LITTLE_ENDIAN ? v : Long.reverseBytes(v)) & (0xffffffffffffffffL >>> ((8 - len) * 8));
           return v;
       }
       return slowReadBytesAsInt64(remaining, len);
   }
   

2. Call this method with len = 0.

What did you expect to see?

When len = 0, I expected the bitwise AND operation (&) with 0xffffffffffffffffL >>> 64 to return a value of 0.

What did you see instead?

In Java 23, the following error is reported:
Shift operation '>>>' by overly large constant value 64

In earlier versions of Java (e.g., 1.8), there is no error, but the result does not match expectations. The operation 0xffffffffffffffffL >>> 64 essentially leaves the result unchanged, where the long value becomes populated with unrelated values, leading to unpredictable outcomes.

In Dart, the same mask (0xffffffffffffffffL >>> 64) would result in 0 as expected.

Anything Else?

This issue could lead to unexpected behavior in different Java versions, and it may require adjustments to the code to handle len = 0 safely across all environments. Please consider revising the implementation to handle the shift operation correctly for len = 0.

Are you willing to submit a PR?

• [x] I'm willing to submit a PR!