indirectbr and phi instructions

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1

Hi,

How does the requirement that phi instructions have one value per predecessor basic block interact with indirectbr instructions? For instance, take the following code:

L1:
br i1 %somevalue, label %L2, label %L3
L2:
%ret1 = i8* blockaddress(@myfunction, %L5)
br label %L4
L3:
%ret2 = i8* blockaddress(@myfunction, %L6)
br label %L4
L4:
%ret = phi i8* [%ret1, L2], [%ret2, L3]
indirectbr i8* %ret, [label %L5, label %L6]
L5:
%myval = phi i32 [0, %L2], [1, %L3] ; are both of these values required, even though the only real possible predecessor block is L2?
ret i32 %myval
L6:
%myval = phi i32 [0, %L2], [1, %L3] ; likewise
ret i32 %myval

Boiled down, I think my question is, “how strict is the ‘one value per predecessor block’ rule on a phi instruction?”

FYI, I am planning on using indirectbr to implement jsr and ret instructions in Java. I fully expect that there are syntax errors in the above code - but I hope you get the idea.

Thanks,

Joshua

2

Actually, if you look at your example, it will fail because of the PHIs in L5 and L6. :slight_smile: The PHI node entries must match up one-to-one and onto with the predecessors. Here's an example that will fail during code-gen:

@a1 = global i8* blockaddress(@foo, %bb4)
@a2 = global i8* blockaddress(@foo, %bb5)

define i32 @foo(i1 %a, i32 %b) {
entry:
  %a1_ = load i8** @a1
  %a2_ = load i8** @a2
  br i1 %a, label %bb1, label %bb2

bb1:
  %v1 = add i32 %b, 37
  br label %bb3

bb2:
  %v2 = add i32 %b, 927
  br label %bb3

bb3:
  %addr = phi i8* [%a1_, %bb1], [%a2_, %bb2]
  indirectbr i8* %addr, [label %bb4, label %bb5]

bb4:
  %v_bb4 = phi i32 [%v1, %bb1], [%v2, %bb2]
  br label %exit

bb5:
  %v_bb5 = phi i32 [%v1, %bb1], [%v2, %bb2]
  br label %exit

exit:
  %p = phi i32 [%v_bb4, %bb4], [%v_bb5, %bb5]
  ret i32 %p
}

-bw

3

Hi Joshua,

L4:
    %ret = phi i8* [%ret1, L2], [%ret2, L3]
    indirectbr i8* %ret, [label %L5, label %L6]
L5:
    %myval = phi i32 [0, %L2], [1, %L3] ; are both of these values
required, even though the only *real* possible predecessor block is L2?

this phi is wrong because it has entries for L2 and L3 which are not
predecessors, while it has no entry for L4 which is a predecessor. The
verifier should catch this.

    ret i32 %myval
L6:
    %myval = phi i32 [0, %L2], [1, %L3] ; likewise
    ret i32 %myval

Likewise.

Ciao,

Duncan.