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cmd/compile: recognize labels even if they have the same name as packages
Another (historic) artifact due to partially resolving symbols too early. Fixes #13539. Change-Id: Ie720c491cfa399599454f384b3a9735e75d4e8f1 Reviewed-on: https://go-review.googlesource.com/17600 Run-TryBot: Robert Griesemer <gri@golang.org> Reviewed-by: Damien Neil <dneil@google.com> TryBot-Result: Gobot Gobot <gobot@golang.org>
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@ -668,10 +668,11 @@ func (p *parser) simple_stmt(labelOk, rangeOk bool) *Node {
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// labelname ':' stmt
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if labelOk {
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// If we have a labelname, it was parsed by operand
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// (calling p.name()) and given an ONAME, ONONAME, or OTYPE node.
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if lhs.Op == ONAME || lhs.Op == ONONAME || lhs.Op == OTYPE {
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// (calling p.name()) and given an ONAME, ONONAME, OTYPE, or OPACK node.
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switch lhs.Op {
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case ONAME, ONONAME, OTYPE, OPACK:
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lhs = newname(lhs.Sym)
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} else {
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default:
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p.syntax_error("expecting semicolon or newline or }")
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// we already progressed, no need to advance
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}
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20
test/fixedbugs/issue13539.go
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20
test/fixedbugs/issue13539.go
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@ -0,0 +1,20 @@
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// errorcheck
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// Copyright 2015 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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// Verify that a label named like a package is recognized
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// as a label rather than a package and that the package
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// remains unused.
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package main
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import "math" // ERROR "imported and not used"
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func main() {
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math:
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for {
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break math
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}
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}
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