How does gcc make optimization about fuction unit hazard and read after
write hazard on a MIPS machine?
I write a simple program:
----------------------
void main (void)
{
int i,s1,s2;
s1=0;
s2=1;
for (i=0;i<100;i++){
s1=s1+i;
s2=s2*i;
}
.....
}
----------------------
use -O2 to compile it, then get the assembly codes like this:
[ 2] 0x10001350: 27 bd ff d0 addiu sp,sp,-48
[ 2] 0x10001354: ff bf 00 20 sd ra,32(sp) // RAW hazard <1>
[ 2] 0x10001358: ff bc 00 18 sd gp,24(sp)
[ 2] 0x1000135c: ff b0 00 10 sd s0,16(sp) // These three `sd'
use the same function unit
[ 2] 0x10001360: 3c 01 00 02 lui at,2
[ 2] 0x10001364: 24 21 b4 24 addiu at,at,-19420 // RAW hazard <2>
[ 2] 0x10001368: 00 39 e0 2d daddu gp,at,t9
[ 5] 0x1000136c: 00 00 28 25 move a1,zero
6] 0x10001370: 24 10 00 01 li s0,1
[ 8] 0x10001374: 00 00 18 25 move v1,zero // RAW hazard <3>
[ 10] 0x10001378: 02 03 00 18 mult s0,v1
[ 10] 0x1000137c: 00 00 80 12 mflo s0
[ 10] 0x10001380: 00 00 00 00 nop
[ 10] 0x10001384: 00 00 00 00 nop
[ 9] 0x10001388: 00 a3 28 21 addu a1,a1,v1
[ 8] 0x1000138c: 24 63 00 01 addiu v1,v1,1
[ 8] 0x10001390: 28 62 00 64 slti v0,v1,100
[ 8] 0x10001394: 14 40 ff f8 bne v0,zero,0x10001378
There are two questions:
1. When we use a superscalar cpu, the above three `sd's are clustered
together and they will be issued at the same time. But why doesn't gcc
schedule them apart ?
2. There are three RAW hazards in the assembly codes. Why doesn't gcc
schedule them far from each other enough to solve the hazards ?
Thank you!
-Junfeng Dong
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