Eliminate array allocation for f(1, *a)

Due to how the compiler works, while f(*a) does not allocate an
array f(1, *a) does.  This is possible to fix in the compiler, but
the change is much more complex.  This attempts to fix the issue
in a simpler way using the peephole optimizer.

Eliminating this array allocation is safe, since just as in the
f(*a) case, nothing else on the caller side can modify the array.

1 files changed, 21 insertions, 0 deletions

diff --git a/compile.c b/compile.c
index 31783feeaf..fe0bd5e6d2 100644
--- a/compile.c
+++ b/compile.c
@@ -3836,6 +3836,27 @@ iseq_peephole_optimize(rb_iseq_t *iseq, LINK_ELEMENT *list, const int do_tailcal
         }
     }
 
+    if (IS_INSN_ID(iobj, splatarray) && OPERAND_AT(iobj, 0) == Qtrue) {
+        LINK_ELEMENT *niobj = &iobj->link;
+
+        /*
+        * Eliminate array allocation for f(1, *a)
+        *
+        *  splatarray true
+        *  send ARGS_SPLAT and not KW_SPLAT|ARGS_BLOCKARG
+        * =>
+        *  splatarray false
+        *  send
+        */
+        if (IS_NEXT_INSN_ID(niobj, send)) {
+            niobj = niobj->next;
+            unsigned int flag = vm_ci_flag((const struct rb_callinfo *)OPERAND_AT(niobj, 0));
+            if ((flag & VM_CALL_ARGS_SPLAT) && !(flag & (VM_CALL_KW_SPLAT|VM_CALL_ARGS_BLOCKARG))) {
+                OPERAND_AT(iobj, 0) = Qfalse;
+            }
+        }
+    }
+
     return COMPILE_OK;
 }