Question

I was scanning a third party source code using Findbugs (just to be cautious before integrating into it mine), and found the following warning:

long a = b << 32 | c

Bug: Integer shift by 32 Pattern id:
ICAST_BAD_SHIFT_AMOUNT, type: BSHIFT,
category: CORRECTNESS

The code performs an integer shift by
a constant amount outside the range
0..31. The effect of this is to use the lower 5 bits of the integer value
to decide how much to shift by. This
probably isn’t want was expected, and
it at least confusing.

Could anyone please explain what exactly does the above mean?

Thanks!
(I am quite a newbie in Java programming)

Answer 1

From the Java Language Specification:

If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f. The shift distance actually used is therefore always in the range 0 to 31, inclusive.

So if b is an int, the expression is identical to

long a = b | c;

which I highly doubt is what is intended. It should probably have been

long a = ((long) b << 32) | c;

(If b is already a long, the code is correct and FindBugs is mistaken about the bug).