Re: [wide-int] int_traits <tree>

[Richard Sandiford <rsandifo at linux dot vnet dot ibm dot com>]

Richard Biener <rguenther@suse.de> writes:
> On Thu, 17 Oct 2013, Richard Sandiford wrote:
>
>> Richard Biener <rguenther@suse.de> writes:
>> >> The new tree representation can have a length greater than max_len
>> >> for an unsigned tree constant that occupies a whole number of HWIs.
>> >> The tree representation of an unsigned 0x8000 is 0x00 0x80 0x00.
>> >> When extended to max_wide_int the representation is the same.
>> >> But a 2-HWI addr_wide_int would be 0x80 0x00, without the leading zero.
>> >> The MIN trims the length from 3 to 2 in the last case.
>> >
>> > Oh, so it was the tree rep that changed?  _Why_ was it changed?
>> > We still cannot use it directly from wide-int and the extra
>> > word is redundant because we have access to TYPE_UNSIGNED (TREE_TYPE ()).
>> 
>> It means that we can now use the tree's HWI array directly, without any
>> copying, for addr_wide_int and max_wide_int.  The only part of decompose ()
>> that does a copy is the small_prec case, which is trivially compiled out
>> for addr_wide_int and max_wide_int.
>
> "     2) addr_wide_int.  This is a fixed size representation that is
>      guaranteed to be large enough to compute any bit or byte sized
>      address calculation on the target.  Currently the value is 64 + 4
>      bits rounded up to the next number even multiple of
>      HOST_BITS_PER_WIDE_INT (but this can be changed when the first
>      port needs more than 64 bits for the size of a pointer).
>
>      This flavor can be used for all address math on the target.  In
>      this representation, the values are sign or zero extended based
>      on their input types to the internal precision.  All math is done
>      in this precision and then the values are truncated to fit in the
>      result type.  Unlike most gimple or rtl intermediate code, it is
>      not useful to perform the address arithmetic at the same
>      precision in which the operands are represented because there has
>      been no effort by the front ends to convert most addressing
>      arithmetic to canonical types.
>
>      In the addr_wide_int, all numbers are represented as signed
>      numbers.  There are enough bits in the internal representation so
>      that no infomation is lost by representing them this way."
>
> so I guess from that that addr_wide_int.get_precision is always
> that "64 + 4 rounded up".  Thus decompose gets that constant precision
> input and the extra zeros make the necessary extension always a no-op.
> Aha.
>
> For max_wide_int the same rules apply, just its size is larger.
>
> Ok.  So the reps are only canonical wide-int because we only
> ever use them with precision > xprecision (maybe we should assert
> that).

No, we allow precision == xprecision for addr_wide_int and max_wide_int too.
But all we do in that case is trim the length.

Requiring precision > xprecision was option (5) from my message.

> Btw, we are not using them directly, but every time we actually
> build a addr_wide_int / max_wide_int we copy them anyway:
>
> /* Initialize the storage from integer X, in precision N.  */
> template <int N>
> template <typename T>
> inline fixed_wide_int_storage <N>::fixed_wide_int_storage (const T &x)
> {
>   /* Check for type compatibility.  We don't want to initialize a
>      fixed-width integer from something like a wide_int.  */
>   WI_BINARY_RESULT (T, FIXED_WIDE_INT (N)) *assertion ATTRIBUTE_UNUSED;
>   wide_int_ref xi (x, N);
>   len = xi.len;
>   for (unsigned int i = 0; i < len; ++i)
>     val[i] = xi.val[i];
> }

Are you saying that:

     max_wide_int x = (tree) y;

should just copy a pointer?  Then what about:

     max_wide_int z = x;

Should that just copy a pointer too, or is it OK for the second constructor
to do a copy?

In most cases we only use the fixed_wide_int to store the result of the
computation.  We don't use the above constructor for things like:

     max_wide_int x = wi::add ((tree) y, (int) z);

wi::add uses y and z without going through max_wide_int.

> What's the reason again to not use my original proposed encoding
> of the MSB being the sign bit?  RTL constants simply are all signed
> then.  Just you have to also sign-extend in functions like lts_p
> as not all constants are sign-extended.  But we can use both tree
> (with the now appended zero) and RTL constants representation
> unchanged.

The answer's the same as always.  RTL constants don't have a sign.
Any time we extend an RTL constant, we need to say whether the extension
should be sign extension or zero extension.  So the natural model for
RTL is that an SImode addition is done to SImode width, not some
arbitrary wider width.

The branch uses wide_int for RTL quite naturally, and changing it
wouldn't help do anything with the tree decompose routine.

Thanks,
Richard