https://gcc.gnu.org/bugzilla/show_bug.cgi?id=92980
--- Comment #11 from Hongyu Wang <wwwhhhyyy333 at gmail dot com> --- (In reply to rguent...@suse.de from comment #10) > > It has two exits which makes it difficult > Or impossible to make it truly do-while. > But it's close enough and further rotating the loop doesn't make it better. Thanks for the explanation. But the original loop without eliminating the parser code is: <bb 2> [local count: 114863532]: goto <bb 5>; [100.00%] <bb 3> [local count: 1014686025]: i.0_1 = (unsigned int) i_10; _2 = i.0_1 * 4; _3 = a_13(D) + _2; _4 = *_3; _5 = i.0_1 + 1; _6 = _5 * 4; _7 = a_13(D) + _6; _8 = *_7; if (_4 > _8) goto <bb 6>; [5.50%] else goto <bb 4>; [94.50%] <bb 4> [local count: 958878293]: i_15 = i_10 + 1; <bb 5> [local count: 1073741824]: # i_10 = PHI <0(2), i_15(4)> _9 = n_12(D) + -1; if (_9 > i_10) goto <bb 3>; [94.50%] else goto <bb 6>; [5.50%] <bb 6> [local count: 114863532]: # _11 = PHI <0(3), 1(5)> return _11; This is considered as not a do-while loop since the latch (bb 4) is not empty, and it is successfully rotated. I don't think there is much difference except i_15 = i_10 + 1 which can be optimized by fre after the parser change. So I wonder if the original one is rotated, why the one with empty latch can't.
