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 <cuong.manhle.vn@gmail.com>
Run-TryBot: Cuong Manh Le <cuong.manhle.vn@gmail.com>
TryBot-Result: Go Bot <gobot@golang.org>
Reviewed-by: Keith Randall <khr@golang.org>
Reviewed-by: Matthew Dempsky <mdempsky@google.com>
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
}
}
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
}
}
--- /dev/null
+// 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...)
+}