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Re: wide-int, C++ front end
- From: Richard Biener <richard dot guenther at gmail dot com>
- To: Jason Merrill <jason at redhat dot com>, Mike Stump <mikestump at comcast dot net>, "gcc-patches at gcc dot gnu dot org Patches" <gcc-patches at gcc dot gnu dot org>, Kenneth Zadeck <zadeck at naturalbridge dot com>, Richard Sandiford <rdsandiford at googlemail dot com>
- Date: Tue, 26 Nov 2013 10:34:53 +0100
- Subject: Re: wide-int, C++ front end
- Authentication-results: sourceware.org; auth=none
- References: <E2C6331E-1CDC-48FC-ADB5-42623433F232 at comcast dot net> <52915DC7 dot 1000200 at redhat dot com> <87vbzg2qo8 dot fsf at talisman dot default>
On Mon, Nov 25, 2013 at 9:05 PM, Richard Sandiford
> Jason Merrill <firstname.lastname@example.org> writes:
>> On 11/23/2013 02:20 PM, Mike Stump wrote:
>>> @@ -2605,8 +2606,7 @@ cp_tree_equal (tree t1, tree t2)
>>> switch (code1)
>>> case INTEGER_CST:
>>> - return TREE_INT_CST_LOW (t1) == TREE_INT_CST_LOW (t2)
>>> - && TREE_INT_CST_HIGH (t1) == TREE_INT_CST_HIGH (t2);
>>> + return wi::to_widest (t1) == wi::to_widest (t2);
>> Why not use wi::eq_p like you do in the C front end?
> Thanks for noticing the difference. I think c_tree_equal should change
> to use to_widest too.
> wi::eq_p (t1, t2) asserts that t1 and t2 are the same precision and
> ignores signedness; it just tests whether they are the same bitstring.
> wi::to_widest (t1) == wi::to_widest (t2) compares them as logical numbers,
> taking sign into account and allowing different types. I think that's
> what the original TREE_INT_CST_LOW and TREE_INT_CST_HIGH tests did too.
Though in this case (comparing two INTEGER_CSTs) it would be better
to use a tree abstraction - thus tree_int_cst_equal. It saves us from
making the decision on what to map this in wide-int to multiple times.
Note that tree_int_cst_equal tests for "same constant value" - in tree
terms this includes sign information and thus requires to_widest.