------- Comment #3 from pinskia at gcc dot gnu dot org 2007-04-10 08:56 ------- IR after assert insertion (plus annotate with ranges found): f (x) { int y; int D.1599;
<bb 2>: if (x_2(D) <= 4) goto <L2>; else goto <L1>; <L1>:; x_7 = ASSERT_EXPR <x_2(D), x_2(D) > 4>; // x_7: [5, +INF] y_4 = x_7 / 4; // y_4: [1, 536870911] # y_1 = PHI <1(2), y_4(3)> // y_1: [1, +INF(OVF)] <L2>:; D.1599_5 = y_1 <= 0; return D.1599_5; } Can someone explain how we go from the union of [1,1] and [1, 536870911] to get an inf with an overflow? Here is another example but this time without a division and just plain obvious no overflow as everything are just compares: int f(int x) { int y; if (x>536870911) y =1; else { if (x<0) x = 1; y = x <= 0; } return y > 2; } -- pinskia at gcc dot gnu dot org changed: What |Removed |Added ---------------------------------------------------------------------------- Severity|normal |blocker Summary|[4.3 Regression] False |[4.3 Regression] False |overflow warning with |overflow warning with |division |division phi nodes http://gcc.gnu.org/bugzilla/show_bug.cgi?id=31522