Hello,

It's a known fact that shifting by more or equal bits as an integer
type contains is undefined behavior per C standard,
http://stackoverflow.com/questions/7214263/unexpected-behavior-of-bitwise-shifting-using-gcc
has relevant quotes and references.

It's less known the rules of dealing with "undefined" values for
compiler writers, in this respect I find LLVM's description insightful:
http://llvm.org/docs/LangRef.html#undefined-values . Summing up, it
says that result of most of the expressions involving undefined value
is also undefined.

Finally, it's known why C standard has it like that: because rules
various CPUs use for such shifts at *runtime* vary, so *runtime* code
should not rely on them. C also doesn't define separation between
compile-time and run-time evaluation strictly, so for it any shift by
too many bits is undefined.

It's all nice and good. But there's difference between "undefined",
"any value" and "weird". Because "too many bits" shifts may be
undefined in C standard, but shifts by arbitrary number of bits are
very well defined in arithmetic - and by very definition of (unsigned)
shift, any value shifted by more bits than available in its
representation is 0. That's logical, that's what users know, that's
what they expect from compiler, and well, that's how x86-gcc behaves.

Now mspgcc:

int v = 5;
int main()
{
    printf("%d\n", v + (1 << 16));
}

main:
        mov     r1, r4
        add     #2, r4
        mov     &v, r15
        add     #1, r15
        push    r15
        push    #.LC0
        call    #printf
        add     #4, r1


So, msp430-gcc just masks out higher bits of shift count, and in this
case leaves original value intact. Which turns term ((1 << BITS) - 1),
which is common to do BITS-modular arithmetic, and would be expected to
just optimize out in case of a full type, into an expression killer with
infinite loops, etc. ensuing.


-- 
Best regards,
 Paul                          mailto:pmis...@gmail.com

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