[EMAIL PROTECTED] (Richard Kenner) wrote on 17/01/2007 18:04:20: > > First, it seems to me that in your case: > > > > (1) a = a | 1 /* a |= 1 */ > > (2) a = a | 1 /* a |= 1 */ > > > > the expressions "a | 1" in (1) and (2) are different as the "a" > > is not the same. So there is nothing to do for CSE. > > It's not a CSE issue, but after (1), you know that the low-order bit of > "a" is a one, so that (2) is a no-op. Note that the similar > a &= ~1; > a &= ~1; > > we do catch in combine. > > It could also be caught by converting > > a = ((a | 1) | 1); > > into > > a = (a | (1 | 1)); >
Yes, you are right if (1) and (2) are in the same basic block. But the initial example that started this thread looks like: a |= 1; if (*x) ... a }= 1; so (1) and (2) are in two separate basic blocks. I think that in this case combine doesn't work. Mircea
