The operand of the shift operator is of type unsigned int.
"x <<= c" is exactly the same as "((int)x) << c"
It doesn't matter whether the promotion is explicit or implicit, the
semantics
are the same.
According to C99, the operand of the shift operator should be "integer
type", which includes "char", "short", "int", "long", or "long long". So
the operand of shift opertor is not necessarily to be unsigned int.
But on the other hand, C99 also requires the "integer promotions are
performed on each of the (bitwise shift operator's) operands", so the
integer promotion should be performed.
----- Original Message -----
From: "Paul Brook" <[EMAIL PROTECTED]>
To: <gcc@gcc.gnu.org>
Cc: "Paul Schlie" <[EMAIL PROTECTED]>
Sent: Sunday, December 17, 2006 10:19 PM
Subject: Re: Char shifts promoted to int. Why?
On Monday 18 December 2006 01:15, Paul Schlie wrote:
Chris Lattner wrote:
> On Dec 17, 2006, at 12:40 PM, Rask Engelmann Lamberts wrote:
>> I seem unable to get a QImode shift instruction from this code:
>>
>> unsigned char x;
>>
>> void qishifttest2 (unsigned int c)
>> {
>> x <<= c;
>> }
>>
>> should have been generated. Also, notice the redundant zero
>> extension.
>> Why are we not generating a QImode shift instruction?
>
> Consider when c = 16. With the (required) integer promotion, the
> result
> is defined (the result is zero). If converted to QImode, the shift
> would
> be undefined, because the (dynamic) shift amount would be larger
> than the
> data type.
??? A left shift >= the precision of its shifted unsigned operand can
only
logically result in a value of 0 regardless of its potential
promotion.
Shifting >= the size of the value being shifted can and do give
nonzero
results on common hardware. Typically hardware will truncate the shift
count.
eg. x << 8 implemented with a QImode shift will give x, not 0.
Although integer promotion as specified by C may technically be
performed
lazily as a function of the implicit target precision required for a
given
operation, GCC tends to initially promote everything and then attempt
to
determine if an operation's precision may be subsequently lowered
after
having already lost critical knowledge of its originally specified
operand's precision.
No. You're confusing some language you just invented with C.
The operand of the shift operator is of type unsigned int.
"x <<= c" is exactly the same as "((int)x) << c"
It doesn't matter whether the promotion is explicit or implicit, the
semantics
are the same.
Paul
