Re: [patch] avoid printing leading 0 in widest_int hex dumps

[Aldy Hernandez <aldyh at redhat dot com>]

On Tue, Oct 17, 2017 at 6:05 PM, Richard Sandiford
<richard.sandiford@linaro.org> wrote:
> Andrew MacLeod <amacleod@redhat.com> writes:
>> On 10/17/2017 08:18 AM, Richard Sandiford wrote:
>>> Aldy Hernandez <aldyh@redhat.com> writes:
>>>> Hi folks!
>>>>
>>>> Calling print_hex() on a widest_int with the most significant bit turned
>>>> on can lead to a leading zero being printed (0x0ffff....). This produces
>>>> confusing dumps to say the least, especially when you incorrectly assume
>>>> an integer is NOT signed :).
>>> That's the intended behaviour though.  wide_int-based types only use as
>>> many HWIs as they need to store their current value, with any other bits
>>> in the value being a sign extension of the top bit.  So if the most
>>> significant HWI in a widest_int is zero, that HWI is there to say that
>>> the previous HWI should be zero- rather than sign-extended.
>>>
>>> So:
>>>
>>>     0x0ffffffff  -> (1 << 32) - 1 to infinite precision
>>>                 (i.e. a positive value)
>>>     0xffffffff   -> -1
>>>
>>> Thanks,
>>> Richard
>>
>> I for one find this very confusing.  If I have a 128 bit value, I don't
>> expect to see a 132 bits.  And there are enough 0's its not obvious when
>> I look.
>
> But Aldy was talking about widest_int, which is wider than 128 bits.
> It's an approximation of infinite precision.

IMO, we should document this leading zero in print_hex, as it's not
inherently obvious.

But yes, I was talking about widest_int.  I should explain what I am
trying to accomplish, since perhaps there is a better way.

I am printing a a wide_int (bounds[i] below), but I really don't want
to print the sign extension nonsense, since it's a detail of the
underlying representation.  Currently I'm doing this to chop off the
unnecessary bits:

/* Wide ints may be sign extended to the full extent of the
   underlying HWI storage, even if the precision we care about
   is smaller.  Chop off the excess bits for prettier output.  */
signop sign = TYPE_UNSIGNED (type) ? UNSIGNED : SIGNED;
widest_int val = widest_int::from (bounds[i], sign);
val &= wi::mask<widest_int> (bounds[i].get_precision (), false);

if (val > 0xffff)
  print_hex (val, pp_buffer (buffer)->digit_buffer);
else
  print_dec (val, pp_buffer (buffer)->digit_buffer, sign);

Since I am calling print_hex() on the widest_int, I get the leading 0.

Do you recommend another way of accomplishing this?

>
> wide_int is the type to use if you want an N-bit number (for some N).
>
>> I don't think a leading 0 should be printed if "precision" bits have
>> already been printed.
>
> Does 0 get printed in that case though?  Aldy's patch skips an upper

No.

Thanks for your input Richard.
Aldy