I am having some issues with how some of the instructions are being legalized.
So this is my intial basic block. The area of concern is the last three instructions. I will pick and choose debug output to keep this small.
SelectionDAG has 36 nodes:
t0: ch = EntryToken
t6: i32,ch = CopyFromReg t0, Register:i32 %vreg507
t2: i32,ch = CopyFromReg t0, Register:i32 %vreg17
t4: i32 = or t2, Constant:i32<256>
t9: i32 = shl t4, Constant:i32<2>
t10: i32 = add t6, t9
t12: i32,ch = CopyFromReg t0, Register:i32 %vreg79
t15: i32,ch = CopyFromReg t0, Register:i32 %vreg1
t16: ch = llvm.tpu.dma.write.1KB.async t0, TargetConstant:i32<4602>, t10, t12, t15
t18: i32,ch = CopyFromReg t0, Register:i32 %vreg166
t20: i32 = AssertZext t18, ValueType:ch:i1
t23: i1 = setcc t20, Constant:i32<0>, seteq:ch
t25: i32,ch = CopyFromReg t0, Register:i32 %vreg396
t28: i1 = setcc t25, Constant:i32<255>, setugt:ch
t29: i1 = and t23, t28
t37: i1 = setcc t29, Constant:i1<-1>, setne:ch
t33: ch = brcond t16, t37, BasicBlock:ch<if.end65.1 0x7097330>
t35: ch = br t33, BasicBlock:ch<if.then64.1 0x7097280>
Here we see that the settcc has been legalized to xor, which I am fine with..
Legalizing: t37: i1 = setcc t29, Constant:i1<-1>, setne:ch
Combining: t37: i1 = setcc t29, Constant:i1<-1>, setne:ch
... into: t38: i1 = xor t29, Constant:i1<-1>
It gets the result from TargetLowering. Does your target override this function?
For some reason I didn’t get Krzysztof Parzyszek e-mail in my inbox, but thanks for the fact I can override the getSetCCResultType. Didn’t even think of that.
But isn’t kinda silly that we transform to xor and then we transform it back. What is the advantage in doing so? Also, since we do that method, I now have to introduce setcc patterns for i1 values, instead of being able to just use logical pattern operators like not.
Legalization transforms the code to only use types and operations that the target declares as natively supported (i.e. "legal"). The DAG combiner, on the other hand tries to simplify the code. The lowering process happens in several stages where the two types of transformations alternate. It may happen that a root node of a certain sub-tree will be transformed from setcc to xor then back to setcc, but the entire sub-tree will change as well.
Why would converting a setcc to xor or vice versa change the sub-trees at all. The inputs to both instructions remain the same. I looked through the codes where both legalization occurs, and it doesn’t seem like any sub-tree modifications should be happening.
I guess the thing is I want to understand why this happens. Is there a reason we want all branch inputs to be setcc operations, instead of a simple xor?
P.S. Still don’t get your messages in my inbox.
I think this is because branch instructions typically depend on a flag in a condition code or flag register. While an instruction like xor writes its output to a normal register. So the setcc reflects an operation that updates flags.
Yes, it seems that nothing else has changed. The presence of xor with both operands being results of setcc could have triggered some transformation in the combiner, but the only outcome that it was able to produce was again setcc with the same operands as the xor. There are a number of restrictions that both, the legalizer and the combiner work under, and it is possible that their changes will in the end amount to something trivial. Make sure that all operations and types that are legal on your target are properly marked as such.