Rotate left 2 bits (like assembler)

[Moneo]

Hey guys, I think we have a confusion here.

The original challenge said: "take a given 16 bit value and do a rotate left 2 bits."

In QB, a 16 bit value implies working with an INTEGER.
If we were working with a LONG, we would be working with 32 bits.

Yes, an INTEGER of -32768 has the same bit pattern as a LONG of 32768, as does an INTEGER of -4 and a LONG of 65532. BUT, both these INTEGERS have the sign bit (bit 15) set to 1, which for an INTEGER means that this is a negative number.

Therefore, if you compared an INTEGER of -32768 to a LONG of 32768, they would not be equal, because the compare is arithmetic, not bit-wise.
If you did an XOR of the INTERGER with the LONG, you would get a zero result showing that the bit patterns are equal because the leading zero bits of the LONG are ignored.

For the testing of the solutions submited, I put the code into a loop, comparing the results against proven results. So, if the submitted code produces answers expressed as LONG, they won't compare with the proven results in INTEGER, when these INTEGERS are negative. That's the problem.

So, for future solutions, please submit code that produces 16 bit values only in integers.
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