removePred and removeArg do different things. removePred moves the last
predecessor to index k, whereas removeArg slides all the args k or
greater down by 1 index.
Kind of unfortunate different behavior in things named similarly.
Fixes #49122
Change-Id: I9ae409bdac744e713f4c121f948e43db6fdc8542
Reviewed-on: https://go-review.googlesource.com/c/go/+/358117
Trust: Keith Randall <khr@golang.org>
Run-TryBot: Keith Randall <khr@golang.org>
TryBot-Result: Go Bot <gobot@golang.org>
Reviewed-by: Cuong Manh Le <cuong.manhle.vn@gmail.com>
if v.Op != OpPhi {
continue
}
- v.RemoveArg(k)
+ n := len(v.Args)
+ v.Args[k].Uses--
+ v.Args[k] = v.Args[n-1]
+ v.Args[n-1] = nil
+ v.Args = v.Args[:n-1]
phielimValue(v)
}
// Fix up child to have one more predecessor.
--- /dev/null
+// compile
+
+// 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 main
+
+var B []bool
+var N int
+
+func f(p bool, m map[bool]bool) bool {
+ var q bool
+ _ = p || N&N < N || B[0] || B[0]
+ return p && q && m[q]
+}