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https://github.com/golang/go
synced 2024-11-23 04:10:04 -07:00
93e3696b5d
When we optimize append(s, make([]T, n)...), we have to be careful not to pass &s[0] + len(s)*sizeof(T) as the argument to memclr, as that pointer might be past-the-end. This can only happen if n is zero, so just special-case n==0 in the generated code. Fixes #67255 Change-Id: Ic680711bb8c38440eba5e759363ef65f5945658b Reviewed-on: https://go-review.googlesource.com/c/go/+/584116 Reviewed-by: Austin Clements <austin@google.com> LUCI-TryBot-Result: Go LUCI <golang-scoped@luci-project-accounts.iam.gserviceaccount.com> Reviewed-by: Cuong Manh Le <cuong.manhle.vn@gmail.com> Reviewed-by: Keith Randall <khr@google.com>
34 lines
922 B
Go
34 lines
922 B
Go
// run
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// Copyright 2024 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 main
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var zero int
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var sink any
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func main() {
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var objs [][]*byte
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for i := 10; i < 200; i++ {
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// The objects we're allocating here are pointer-ful. Some will
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// max out their size class, which are the ones we want.
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// We also allocate from small to large, so that the object which
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// maxes out its size class is the last one allocated in that class.
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// This allocation pattern leaves the next object in the class
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// unallocated, which we need to reproduce the bug.
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objs = append(objs, make([]*byte, i))
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}
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sink = objs // force heap allocation
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// Bug will happen as soon as the write barrier turns on.
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for range 10000 {
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sink = make([]*byte, 1024)
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for _, s := range objs {
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s = append(s, make([]*byte, zero)...)
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}
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}
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}
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