cmd/compile: fix identical to recognize any and interface{}

Currently, identical handles any and interface{} by checking against
Types[TINTER]. This is not always true, since when two generated
interface{} types may not use the same *Type instance.

Instead, we must check whether Type is empty interface or not.

Fixes #49875

Change-Id: I28fe4fc0100041a01bb03da795cfe8232b515fc4
Reviewed-on: https://go-review.googlesource.com/c/go/+/367754
Trust: Cuong Manh Le <[email protected]>
Run-TryBot: Cuong Manh Le <[email protected]>
TryBot-Result: Go Bot <[email protected]>
Reviewed-by: Keith Randall <[email protected]>
Reviewed-by: Matthew Dempsky <[email protected]>

src/cmd/compile/internal/types/identity.go

@@ -59,7 +59,8 @@ func identical(t1, t2 *Type, flags int, assumedEqual map[typePair]struct{}) bool
 		case TINT32:
 			return (t1 == Types[TINT32] || t1 == RuneType) && (t2 == Types[TINT32] || t2 == RuneType)
 		case TINTER:
-			return (t1 == Types[TINTER] || t1 == AnyType) && (t2 == Types[TINTER] || t2 == AnyType)
+			// Make sure named any type matches any empty interface.
+			return t1 == AnyType && t2.IsEmptyInterface() || t2 == AnyType && t1.IsEmptyInterface()
 		default:
 			return false
 		}

src/cmd/compile/internal/types/type.go

@@ -1211,7 +1211,8 @@ func (t *Type) cmp(x *Type) Cmp {
 			}
 
 		case TINTER:
-			if (t == Types[AnyType.kind] || t == AnyType) && (x == Types[AnyType.kind] || x == AnyType) {
+			// Make sure named any type matches any empty interface.
+			if t == AnyType && x.IsEmptyInterface() || x == AnyType && t.IsEmptyInterface() {
 				return CMPeq
 			}
 		}

test/typeparam/issue49875.go

@@ -0,0 +1,14 @@
+// compile -G=3
+
+// Copyright 2021 The Go Authors. All rights reserved.
+// Use of this source code is governed by a BSD-style
+// license that can be found in the LICENSE file.
+
+package p
+
+func f(args ...interface{}) {}
+
+func g() {
+	var args []any
+	f(args...)
+}