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From: "glisse at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org> To: gcc-bugs@gcc.gnu.org Subject: [Bug middle-end/64309] if (1 & (1 << n)) not simplified to if (n == 0) Date: Mon, 15 Dec 2014 20:19:00 -0000 [thread overview] Message-ID: <bug-64309-4-42RBkRefFd@http.gcc.gnu.org/bugzilla/> (raw) In-Reply-To: <bug-64309-4@http.gcc.gnu.org/bugzilla/> https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64309 --- Comment #6 from Marc Glisse <glisse at gcc dot gnu.org> --- (In reply to Marek Polacek from comment #5) > I don't think so. I tried to come up with a more general transformation > that would simplify ((CST << n) & CST) != 0, but I haven't found anything If both CST are (possibly different) powers of 2 it works. ((c<<n)&c)==c also works. ((pow2<<p)&(pow2<<n))!=0. > yet. So maybe just this? > ((1 << n) & 1) != 0 -> n == 0 That looks like what richi posted. For scalars, != 0 is useless and this could just be: (1<<n)&1 -> n==0 > ((1 << n) & 1) == 0 -> n != 0
next prev parent reply other threads:[~2014-12-15 20:19 UTC|newest] Thread overview: 15+ messages / expand[flat|nested] mbox.gz Atom feed top 2014-12-15 2:49 [Bug rtl-optimization/64309] New: " olegendo at gcc dot gnu.org 2014-12-15 11:45 ` [Bug middle-end/64309] " mpolacek at gcc dot gnu.org 2014-12-15 13:23 ` mpolacek at gcc dot gnu.org 2014-12-15 19:08 ` olegendo at gcc dot gnu.org 2014-12-15 19:22 ` mpolacek at gcc dot gnu.org 2014-12-15 20:19 ` glisse at gcc dot gnu.org [this message] 2014-12-15 20:41 ` olegendo at gcc dot gnu.org 2014-12-15 20:46 ` jakub at gcc dot gnu.org 2014-12-15 20:46 ` mpolacek at gcc dot gnu.org 2014-12-15 20:49 ` mpolacek at gcc dot gnu.org 2014-12-15 21:06 ` glisse at gcc dot gnu.org 2014-12-15 21:11 ` glisse at gcc dot gnu.org 2014-12-16 8:54 ` rguenther at suse dot de 2014-12-16 18:29 ` mpolacek at gcc dot gnu.org 2014-12-16 18:30 ` mpolacek at gcc dot gnu.org
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