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go/test/stack.go
Russ Cox 1fa4173444 5l, 8l: pass stack frame size to morestack when needed
Shame on me: I fixed the same bug in 6l in 8691fcc6a66e
(https://golang.org/cl/2609041) and neglected
to look at 5l and 8l to see if they were affected.

On the positive side, the check I added in that CL is the
one that detected this bug.

Fixes #1457.

R=ken2
CC=golang-dev
https://golang.org/cl/3981052
2011-02-01 18:34:41 -05:00

100 lines
1.7 KiB
Go

// $G $D/$F.go && $L $F.$A && ./$A.out
// Copyright 2009 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.
// Try to tickle stack splitting bugs by doing
// go, defer, and closure calls at different stack depths.
package main
type T [20]int
func g(c chan int, t T) {
s := 0
for i := 0; i < len(t); i++ {
s += t[i]
}
c <- s
}
func d(t T) {
s := 0
for i := 0; i < len(t); i++ {
s += t[i]
}
if s != len(t) {
println("bad defer", s)
panic("fail")
}
}
func f0() {
// likely to make a new stack for f0,
// because the call to f1 puts 3000 bytes
// in our frame.
f1()
}
func f1() [3000]byte {
// likely to make a new stack for f1,
// because 3000 bytes were used by f0
// and we need 3000 more for the call
// to f2. if the call to morestack in f1
// does not pass the frame size, the new
// stack (default size 5k) will not be big
// enough for the frame, and the morestack
// check in f2 will die, if we get that far
// without faulting.
f2()
return [3000]byte{}
}
func f2() [3000]byte {
// just take up space
return [3000]byte{}
}
var c = make(chan int)
var t T
var b = []byte{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
func recur(n int) {
ss := string(b)
if len(ss) != len(b) {
panic("bad []byte -> string")
}
go g(c, t)
f0()
s := <-c
if s != len(t) {
println("bad go", s)
panic("fail")
}
f := func(t T) int {
s := 0
for i := 0; i < len(t); i++ {
s += t[i]
}
s += n
return s
}
s = f(t)
if s != len(t)+n {
println("bad func", s, "at level", n)
panic("fail")
}
if n > 0 {
recur(n - 1)
}
defer d(t)
}
func main() {
for i := 0; i < len(t); i++ {
t[i] = 1
}
recur(8000)
}