Hi,
I am investigating a test that is failing. I noticed that the
linearized output for:
static double cbrtl (double x)
{
int hx;
double r,s,w;
double lt;
unsigned sign;
union {
double t;
unsigned pt[2];
} ut, ux;
int n0;
ut.t = 1.0;
n0 = (ut.pt[0] == 0);
ut.t = 0.0;
ux.t = x;
is this:
cbrtl:
.L0:
<entry-point>
load.32 %r2 <- 0[ut]
seteq.32 %r3 <- %r2, $0
set.64 %r4 <- 0.000000
store.64 %r4 -> 0[ut]
store.64 %arg1 -> 0[ux]
The assignment of 1.0 to ut.t has been skipped. Does this look
correct? Is there a way to switch off optimisation / simplification?
Thanks and Regards
Dibyendu
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