[Bug tree-optimization/53804] New: branch reordering missed optimization
steven at gcc dot gnu.org
gcc-bugzilla@gcc.gnu.org
Fri Jun 29 10:36:00 GMT 2012
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=53804
Bug #: 53804
Summary: branch reordering missed optimization
Classification: Unclassified
Product: gcc
Version: unknown
Status: UNCONFIRMED
Severity: enhancement
Priority: P3
Component: tree-optimization
AssignedTo: unassigned@gcc.gnu.org
ReportedBy: steven@gcc.gnu.org
Consider this test case:
int
foo1 (int a, int b)
{
if (a > 0)
return 1;
else if (b > 0 && a < 0)
return -3;
return 9;
}
int
foo2 (int a, int b)
{
if (a > 0)
return 1;
else if (a < 0 && b > 0)
return -3;
return 9;
}
Ideally these two functions would be optimized to the same code, because they
are semantically equivalent. The ideal form is foo2 because the result of the
first comparison against "a" can be re-used for the second test, but GCC does
not perform this optimization. The .227r.final dump looks like this on
powerpc64-unknown-linux-gnu (all notes removed for readability):
;; Function foo1 (foo1, funcdef_no=0, decl_uid=1997, cgraph_uid=0)
11 %7:CC=cmp(%3:SI,0) # r7 = cmp(a,0)
5 %9:DI=0x1 # r9 = 1
12 pc={(%7:CC<=0)?L74:pc} # if (r7 <= 0) goto L74
L20:
26 %3:DI=%9:DI # r3 = r9
29 use %3:DI # ..
64 return # return r3
i 63: barrier
L74:
14 %7:CC=cmp(%4:SI,0) # r7 = cmp (b,0)
8 %9:DI=0x9 # r9 = 9
15 pc={(%7:CC<=0)?L20:pc} # if (r7 <= 0) goto L20
53 %9:DI=-%3:DI==0 # r9 = -(r3==0)
54 {%9:DI=%9:DI&0xc;clobber scratch;} # r9 = r9 & 12
55 %9:DI=%9:DI-0x3 # r9 = r9 - 3
68 %3:DI=%9:DI # r3 = r9
69 use %3:DI # ..
70 return # return r3
i 73: barrier
;; Function foo2 (foo2, funcdef_no=1, decl_uid=2001, cgraph_uid=1)
11 %7:CC=cmp(%3:SI,0) # r7 = cmp(a,0)
12 pc={(%7:CC<=0)?L57:pc} # if (r7 <= 0) goto L57
5 %3:DI=0x1 # r3 = 1
29 use %3:DI # ..
56 return # return r3
i 55: barrier
L57:
14 %7:CC=cmp(%3:DI,0) # r7 = cmp(a,0) // ??? redundant
8 %3:DI=0x9 # r3 = 9
51 use %3:DI # ..
15 pc={(%7:CC==0)?return:pc} # if (r7 == 0) return r3
17 %7:CC=cmp(%4:SI,0) # r7 = cmp(b,0)
52 use %3:DI # ..
18 pc={(%7:CC<=0)?return:pc} # if (r7 <= 0) return r3
6 %3:DI=0xfffffffffffffffd # r3 = -3
53 use %3:DI # ..
54 return # return r3
i 47: barrier
Note how foo1 needs two branches whereas foo2 only needs 1.
(I'm not sure why there is the redundant compare in foo2:insn 14)
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