https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752
--- Comment #18 from rguenther at suse dot de <rguenther at suse dot de> --- On Tue, 19 May 2015, gil.hur at sf dot snu.ac.kr wrote: > https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65752 > > --- Comment #17 from Chung-Kil Hur <gil.hur at sf dot snu.ac.kr> --- > Hi Richard, > > I modified the example further. > > #include <stdio.h> > > int main() { > int x = 0; > uintptr_t xp = (uintptr_t) &x; > uintptr_t i, j; > > for (i = 0; i < xp; i++) { } > j = i; > /* The following "if" statement is never executed because j == xp */ > if (j != xp) { > printf("hello\n"); > j = xp; > } Here j is always xp and thus ... > *(int*)((xp+i)-j) = 15; ... this can (and is) simplified to *(int *)i = 15; making it the same testcase again. > printf("%d\n", x); > } > > The above example does not print "hello", so i can assume that "j = xp" is not > executed. > However, the program prints "0" instead of "15". > Can you explain this?
