[LLVMdev] Terminator found in the middle of a basic block error

J. Charles N. M. jcharles.nmbiada at gmail.com
Mon May 5 07:43:28 PDT 2014


Hello,
I am currently building a compilator for an extended CIL to LLVM with the
OCaml's LLVM API.

For all the branch statements, I need to create some basic block to
represente the LLVM branchement.
But after compilation, I got a module error

See message error below :

Terminator found in the middle of a basic block!
label %lbl3
Broken module found, compilation aborted!
Aborted (core dumped)

-------
And here is the C source code follows by its compilation in LLVM IR
representation.

C Source:
int f(void)
{
  int c;
  c = 170;
  if (c % 7 > 5) c = 7;
  return c;
}

int main(void)
{
  int tmp;
  tmp = f();
  return tmp;
}

LLVM module :
; ModuleID = 'program'

define i32 @f() {
entry:
  %c = alloca i32
  store i32 170, i32* %c
  %fclv2 = load i32* %c
  %fclv3 = srem i32 %fclv2, 7
  %fclv4 = icmp sgt i32 %fclv3, 5
  br i1 %fclv4, label %lbl1, label %lbl2

lbl1:                                             ; preds = %entry
  store i32 7, i32* %c
  br label %lbl3

lbl2:                                             ; preds = %entry
  br label %lbl3

lbl3:                                             ; preds = %lbl2, %lbl1
  store i32 7, i32* %c
  %fclv5 = load i32* %c
  ret i32 %fclv5
}

Until this optimized form does not work!

define i32 @f() {
entry:
  %c = alloca i32
  store i32 170, i32* %c
  %fclv2 = load i32* %c
  %fclv3 = srem i32 %fclv2, 7
  %fclv4 = icmp sgt i32 %fclv3, 5
  br i1 %fclv4, label %lbl1, label %lbl3

lbl1:                                             ; preds = %entry
  store i32 7, i32* %c
  br label %lbl3

lbl3:                                             ; preds = %entry, %lbl1
  store i32 7, i32* %c
  %fclv5 = load i32* %c
  ret i32 %fclv5
}

I cannot understand why my module is corrupted.
Reference  manuel says
"[...] every basic block in a program ends with a “Terminator” instruction,
which indicates which block should be executed after the current block is
finished. These terminator instructions typically yield a ‘void‘ value:
they produce control flow, not values (the one exception being the ‘invoke‘
instruction).
The terminator instructions are: ‘ret‘, ‘br‘, ‘switch‘, ‘indirectbr‘,
‘invoke‘, ‘resume‘, and ‘unreachable‘."

So whats happened here ?
Thanks for your helps.
 ---
| J. Charles
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