Re: logical error with 16bit arithmatic operations

[Ian Lance Taylor <iant at google dot com>]

daniel tian <daniel.xntian@gmail.com> writes:

>       I am porting gcc to a 32bit RISC chip, and  I met a logical
> error with 16bit arithmetic operations in generating assemble code.
> the error is between two 16bit data movement(unsigned short).
> While like A = B,  A, and B are all unsigned short type.  B is a
> result of a series of computation, which may have value in high 16bit.
> Because A , B are both HImode, so the move insn doesn't take zero
> extend operation, so A will have value in high 16bit which will caused
> a wrong result in computation.

The relevant types in your code are "unsigned short".  I assume that on
your processor that is a 16 bit type.  An variable of an unsigned 16 bit
type can hold values from 0 to 0xffff, inclusive.  There is nothing
wrong with having the high bit be set in such a variable.  It just means
that the value is between 0x8000 and 0xffff.


> ;;	  1--> 23   R5=R6 0>>0xc                       :nothing
>              //R5 and R6 are both unsigned short type. But R6 will
> have it's value in high 16bit because of series of logical operations
> (AND, OR, XOR and so on). Doing it like this way will cause R5 also
> being valued in high 16bit. This will cause a wrong value. The correct
> one is : R5 = R2 0 >> 0xC. because R2 did zero extend in former insn
>>From R6. But How I let gcc know it.

No.  0>> means a logical right shift, not an arithmetic right shift
(LSHIFTRT rathre than ASHIFTRT).  When doing a logical right shift, the
sign bit is moved right also; it is not sticky.  This instruction is
fine.  Perhaps there is a problem in your implementation of LSHIFTRT.

Ian