After having called linearize_fn_statement() the active BB
is tested but at this point there is always an active BB as
linearize_fn_statement() always create one.
So, remove this unneeded test.
Signed-off-by: Luc Van Oostenryck <luc.vanoostenryck@xxxxxxxxx>
---
linearize.c | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/linearize.c b/linearize.c
index 9cbfeb786..6570c42d1 100644
--- a/linearize.c
+++ b/linearize.c
@@ -2016,7 +2016,7 @@ static pseudo_t linearize_inlined_call(struct entrypoint *ep, struct statement *
use_pseudo(insn, symbol_pseudo(ep, stmt->inline_fn), &insn->func);
bb = ep->active;
- if (bb && !bb->insns)
+ if (!bb->insns)
bb->pos = stmt->pos;
add_one_insn(ep, insn);
return pseudo;
@@ -2458,7 +2458,7 @@ static struct entrypoint *linearize_fn(struct symbol *sym, struct symbol *base_t
} END_FOR_EACH_PTR(arg);
result = linearize_fn_statement(ep, stmt);
- if (bb_reachable(ep->active) && !bb_terminated(ep->active)) {
+ if (!bb_terminated(ep->active)) {
struct symbol *ret_type = base_type->ctype.base_type;
struct instruction *insn = alloc_typed_instruction(OP_RET, ret_type);
--
2.18.0
