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cmd/cgo: don't replace newlines with semicolons in expressions
Fixes #29781 Change-Id: Id032d07a54b8c24f0c6d3f6e512932f76920ee04 Reviewed-on: https://go-review.googlesource.com/c/158457 Run-TryBot: Ian Lance Taylor <iant@golang.org> TryBot-Result: Gobot Gobot <gobot@golang.org> Reviewed-by: Brad Fitzpatrick <bradfitz@golang.org>
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17
misc/cgo/test/issue29781.go
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17
misc/cgo/test/issue29781.go
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@ -0,0 +1,17 @@
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// Copyright 2019 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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// Error with newline inserted into constant expression.
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// Compilation test only, nothing to run.
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package cgotest
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// static void issue29781F(char **p, int n) {}
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// #define ISSUE29781C 0
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import "C"
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func issue29781G() {
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var p *C.char
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C.issue29781F(&p, C.ISSUE29781C+1)
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}
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@ -131,12 +131,27 @@ func gofmt(n interface{}) string {
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// AST expression onto a single line. The lexer normally inserts a
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// semicolon at each newline, so we can replace newline with semicolon.
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// However, we can't do that in cases where the lexer would not insert
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// a semicolon. Fortunately we only have to worry about cases that
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// can occur in an expression passed through gofmt, which just means
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// composite literals.
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// a semicolon. We only have to worry about cases that can occur in an
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// expression passed through gofmt, which means composite literals and
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// (due to the printer possibly inserting newlines because of position
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// information) operators.
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var gofmtLineReplacer = strings.NewReplacer(
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"{\n", "{",
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",\n", ",",
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"++\n", "++;",
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"--\n", "--;",
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"+\n", "+",
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"-\n", "-",
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"*\n", "*",
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"/\n", "/",
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"%\n", "%",
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"&\n", "&",
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"|\n", "|",
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"^\n", "^",
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"<\n", "<",
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">\n", ">",
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"=\n", "=",
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",\n", ",",
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"\n", ";",
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)
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