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Another performance regression

Try the program at the bottom with -O2 -funroll-loops. Don't worry about the body
of the loops; that's only important insofar as it has the right amount of code
to cause the inner loop to be unrolled the right number of times, namely 2, with
1 left over. The unroller generates some rather stupid code here:

/* Calculate the difference between the final and initial values.
Final value may be a (plus (reg x) (const_int 1)) rtx.
Let the following cse pass simplify this if initial value is
a constant.

(there's more to it besides the expression described above)
with the expectation that cse will clean it up. However, the second pass of
loop optimization pulls some, but not all, of this code out of the
outer loop, with the effect that cse can't eliminate it. On ppc, for example,
the beginning of the function looks like this:

bge- cr0,L18 ; zero-trip check for outer loop
li r0,1 ; unnecessary
cmpwi cr1,r0,0 ; unnecessary
cmpwi cr6,r0,25 ; unnecessary
L16: ; top of outer loop
slwi r0,r6,2
li r8,0
add r7,r0,r28
mr r10,r29
bge+ cr6,L22 ; always false
beq- cr1,L15 ; always false
... single copy of inner loop body...
... two copies of inner loop body, executed 12 times...
ble L15
blt L16

I'm not entirely sure, but I think this patch was the culprit:

2002-07-21 Richard Henderson <>

* loop.h (LOOP_AUTO_UNROLL): Rename from LOOP_FIRST_PASS.
* loop.c (strength_reduce): Update.
* toplev.c (rest_of_compilation): Do unrolling in the first
loop pass, not the second.

This didn't happen when unrolling was done last.

So should I fix this by making the unrolling code smarter, in effect
doing cse's job? It seems likely Roger Sayle's approach of running
gcse after loop opts would Just Work. Is that going to go in?

int foo(char *abcd00, int abcd01, char *abcd02, int *abcd03, int*abcd04) {
int abcd05, abcd06, abcd07=0, abcd08=0, abcd09, abcd10, abcd11=0;
for (abcd05=0;abcd05<abcd01;abcd05++) {
for(abcd06=0;abcd06<25;abcd06++) {
if(abcd00[abcd05]==abcd06 && abcd07<2) {
if (abcd07==0) {
abcd09=26*abcd03[abcd06]; abcd02[abcd08++]=abcd00[abcd05]; abcd07=1;
} else if (abcd07==1) {
abcd10=abcd09+abcd03[abcd06]; abcd02[abcd08++]=abcd00[abcd05]; abcd07=2;
if(abcd07==2) {
abcd04[abcd11++]=abcd10; abcd07=0;
return abcd07;

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