Andrew Poelstra wrote On 06/02/06 14:59,:
> On 2006-06-02, Eric Sosman <> wrote:
>
>>
>>Andrew Poelstra wrote On 06/02/06 11:26,:
>>
>>>On 2006-06-02, Richard Bos <> wrote:
>>>
>>>
>>>>"Tomás" <No.Email@Address> wrote:
>>>>
>>>>
>>>>
>>>>>What's the best portable way to set the MSB of an unsigned integer type?
>>>>>
>>>>>Is the following any good?:
>>>>>
>>>>> Start of with: 0000 0000
>>>>> Flip all the bits: 1111 1111
>>>>> Shift once to the right: 0111 1111
>>>>> Flip all the bits: 1000 0000
>>>>>
>>>>>Here it is done in code:
>>>>>
>>>>> typedef unsigned short UIntType;
>>>>>
>>>>> UIntType msb_only =
>>>>> ~( ~( (UIntType)0 ) >> 1 );
>>>>>
>>>>>Is there a better way?
>>>>
>>>>Because of the way unsigned integer overflow is handled in C, you can
>>>>replace the first two steps with converting -1 to the desired type. This
>>>>results in
>>>>
>>>> UIntType msb_only = ~( (UIntType)-1 >> 1 );
>>>>
>>>>This is obviously shorter; up to you to decide whether you find it as
>>>>legible.
>>>
>>>
>>>I would take one and left-shift it sizeof(type) * CHAR_BIT.
>>>
>>>This solution is pretty easy to read:
>>>
>>>int m = 1 << (sizeof m * CHAR_BIT) /* Set MSB */
>>
>> This is wrong. R-O-N-G, wrong. Where to begin?
>>
>> - It assumes all bits of an `int' are value bits, and
>> ignores the possibility of padding bits. All right,
>> that may be more of a "theoretical" than an "actual"
>> problem, but it's not the only one ...
>>
>> - Shift operators are only defined when the shift
>> distance is strictly less than the width of the
>> shifted value. There's a `-1' missing, without which
>> the above yields undefined behavior (6.5.7/3). On
>> actual machines, the likely result is `m=0' or `m=1'.
>>
>> - Even with the `-1', the above is an attempt to shift
>> a value bit into the sign position, which once again
>> yields undefined behavior (6.5.7/4). The missing `-1'
>> should perhaps be a `-2', depending on how you choose
>> to define the "M"SB of a signed integer.
>>
>> - Speaking of signed integers, the O.P. specifically
>> asked about *un*signed integers.
>>
>> If somebody offers you this "solution," I'd recommend
>>that you not drink it.
>>
>
>
> Point 1 is generally not a concern for primitive types.
> Point 2 is correct; I did forget a -1.
> Point 3 is eliminated by point 4, and in fact.
>
> My definition of MSB is leftmost bit. My solution (with a -1)
> works by that definition.
Well, it works once you've changed from `int' to
`unsigned int' (in *two* places) and tacked on a `-1',
provided there are no padding bits. Putting all this
together and generalizing to types that might be wider
than an `int', you wind up with
UIntType m = (UIntType)1 << (CHAR_BIT * sizeof m - 1);
Readability is in the eye of the beholder, but I don't
find this more readable than
UIntType m = ((UIntType)-1 >> 1) + 1;
... which has the virtues of being both bullet-proof and
shorter. This beholder's eye sees no reason to prefer
the longer, shakier construct.
By the way, note that `~((UIntType)-1 >> 1)' is not
guaranteed to work. If UIntType is sufficiently narrow it
will be subject to the "integer promotions" and the value
inside the parentheses will be a non-negative signed `int'
(non-negative because otherwise promotion wouldn't have
occurred). Applying `~' yields a non-positive value, but
just what that value is depends on how the system represents
negative integers. On a ones' complement machine, converting
back to UIntType would give an unintended result.
--
Similar Threads
