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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 <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>
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@ -59,7 +59,8 @@ func identical(t1, t2 *Type, flags int, assumedEqual map[typePair]struct{}) bool
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case TINT32:
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return (t1 == Types[TINT32] || t1 == RuneType) && (t2 == Types[TINT32] || t2 == RuneType)
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case TINTER:
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return (t1 == Types[TINTER] || t1 == AnyType) && (t2 == Types[TINTER] || t2 == AnyType)
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// Make sure named any type matches any empty interface.
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return t1 == AnyType && t2.IsEmptyInterface() || t2 == AnyType && t1.IsEmptyInterface()
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default:
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return false
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}
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@ -1211,7 +1211,8 @@ func (t *Type) cmp(x *Type) Cmp {
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}
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case TINTER:
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if (t == Types[AnyType.kind] || t == AnyType) && (x == Types[AnyType.kind] || x == AnyType) {
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// Make sure named any type matches any empty interface.
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if t == AnyType && x.IsEmptyInterface() || x == AnyType && t.IsEmptyInterface() {
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return CMPeq
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}
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}
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14
test/typeparam/issue49875.go
Normal file
14
test/typeparam/issue49875.go
Normal file
@ -0,0 +1,14 @@
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// compile -G=3
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// Copyright 2021 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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package p
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func f(args ...interface{}) {}
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func g() {
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var args []any
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f(args...)
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}
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