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Add a general purpose unbounded buffer implementation #3099
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,78 @@ | ||
/* | ||
* Copyright 2019 gRPC authors. | ||
* | ||
* Licensed under the Apache License, Version 2.0 (the "License"); | ||
* you may not use this file except in compliance with the License. | ||
* You may obtain a copy of the License at | ||
* | ||
* http://www.apache.org/licenses/LICENSE-2.0 | ||
* | ||
* Unless required by applicable law or agreed to in writing, software | ||
* distributed under the License is distributed on an "AS IS" BASIS, | ||
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. | ||
* See the License for the specific language governing permissions and | ||
* limitations under the License. | ||
* | ||
*/ | ||
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||
// Package buffer provides an implementation of an unbounded buffer. | ||
package buffer | ||
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import "sync" | ||
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// Unbounded is an implementation of an unbounded buffer which does not use | ||
// extra goroutines. This is typically used for passing updates from one entity | ||
// to another within gRPC. | ||
// | ||
// All methods on this type are thread-safe and don't block on anything except | ||
// the underlying mutex used for synchronization. | ||
type Unbounded struct { | ||
c chan interface{} | ||
mu sync.Mutex | ||
backlog []interface{} | ||
} | ||
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// NewUnbounded returns a new instance of Unbounded. | ||
func NewUnbounded() *Unbounded { | ||
return &Unbounded{c: make(chan interface{}, 1)} | ||
} | ||
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// Put adds t to the unbounded buffer. | ||
func (b *Unbounded) Put(t interface{}) { | ||
b.mu.Lock() | ||
if len(b.backlog) == 0 { | ||
select { | ||
case b.c <- t: | ||
b.mu.Unlock() | ||
return | ||
default: | ||
} | ||
} | ||
b.backlog = append(b.backlog, t) | ||
b.mu.Unlock() | ||
} | ||
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// Load sends the earliest buffered data, if any, onto the read channel | ||
// returned by Get(). Users are expected to call this every time they read a | ||
// value from the read channel. | ||
func (b *Unbounded) Load() { | ||
b.mu.Lock() | ||
if len(b.backlog) > 0 { | ||
select { | ||
case b.c <- b.backlog[0]: | ||
b.backlog[0] = nil | ||
b.backlog = b.backlog[1:] | ||
default: | ||
} | ||
} | ||
b.mu.Unlock() | ||
} | ||
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// Get returns a read channel on which values added to the buffer, via Put(), | ||
// are sent on. | ||
// | ||
// Upon reading a value from this channel, users are expected to call Load() to | ||
// send the next buffered value onto the channel if there is any. | ||
func (b *Unbounded) Get() <-chan interface{} { | ||
return b.c | ||
} |
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@@ -0,0 +1,111 @@ | ||
/* | ||
* Copyright 2019 gRPC authors. | ||
* | ||
* Licensed under the Apache License, Version 2.0 (the "License"); | ||
* you may not use this file except in compliance with the License. | ||
* You may obtain a copy of the License at | ||
* | ||
* http://www.apache.org/licenses/LICENSE-2.0 | ||
* | ||
* Unless required by applicable law or agreed to in writing, software | ||
* distributed under the License is distributed on an "AS IS" BASIS, | ||
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. | ||
* See the License for the specific language governing permissions and | ||
* limitations under the License. | ||
* | ||
*/ | ||
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package buffer | ||
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import ( | ||
"reflect" | ||
"sort" | ||
"sync" | ||
"testing" | ||
) | ||
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const ( | ||
numWriters = 10 | ||
numWrites = 10 | ||
) | ||
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// wantReads contains the set of values expected to be read by the reader | ||
// goroutine in the tests. | ||
var wantReads []int | ||
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func init() { | ||
for i := 0; i < numWriters; i++ { | ||
for j := 0; j < numWrites; j++ { | ||
wantReads = append(wantReads, i) | ||
} | ||
} | ||
} | ||
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// TestSingleWriter starts one reader and one writer goroutine and makes sure | ||
// that the reader gets all the value added to the buffer by the writer. | ||
func TestSingleWriter(t *testing.T) { | ||
ub := NewUnbounded() | ||
reads := []int{} | ||
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var wg sync.WaitGroup | ||
wg.Add(1) | ||
go func() { | ||
defer wg.Done() | ||
ch := ub.Get() | ||
for i := 0; i < numWriters*numWrites; i++ { | ||
r := <-ch | ||
reads = append(reads, r.(int)) | ||
ub.Load() | ||
} | ||
}() | ||
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wg.Add(1) | ||
go func() { | ||
defer wg.Done() | ||
for i := 0; i < numWriters; i++ { | ||
for j := 0; j < numWrites; j++ { | ||
ub.Put(i) | ||
} | ||
} | ||
}() | ||
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wg.Wait() | ||
if !reflect.DeepEqual(reads, wantReads) { | ||
t.Errorf("reads: %#v, wantReads: %#v", reads, wantReads) | ||
} | ||
} | ||
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// TestMultipleWriters starts multiple writers and one reader goroutine and | ||
// makes sure that the reader gets all the data written by all writers. | ||
func TestMultipleWriters(t *testing.T) { | ||
ub := NewUnbounded() | ||
reads := []int{} | ||
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var wg sync.WaitGroup | ||
wg.Add(1) | ||
go func() { | ||
defer wg.Done() | ||
ch := ub.Get() | ||
for i := 0; i < numWriters*numWrites; i++ { | ||
r := <-ch | ||
reads = append(reads, r.(int)) | ||
ub.Load() | ||
} | ||
}() | ||
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wg.Add(numWriters) | ||
for i := 0; i < numWriters; i++ { | ||
go func(index int) { | ||
defer wg.Done() | ||
for j := 0; j < numWrites; j++ { | ||
ub.Put(index) | ||
} | ||
}(i) | ||
} | ||
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wg.Wait() | ||
sort.Ints(reads) | ||
if !reflect.DeepEqual(reads, wantReads) { | ||
t.Errorf("reads: %#v, wantReads: %#v", reads, wantReads) | ||
} | ||
} |
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Is the buffer ordered? If Put(A) happens before Put(C), is Get guaranteed to return A before C? If so, consider this series of events where goroutines 1 and 2 are executing concurrently.
Goroutine 1:
1.1. Put(A)
1.2. Put(B)
1.3. Get()
1.4. Load()
Goroutine 2:
2.1. Put(C)
Between 1.3 and 1.4, let's say 2.1 happens. Since
<- b.Get
has completed, the chan is empty, so 2.1 puts C into the chan. Now the channel has C, but the backlog has B, so C would be returned before B the next time Get() is called even though B was inserted before C. (1.4's execution doesn't matter anymore because Load's select executes the empty default and simply returns.)If this isn't ordered, ignore everything I said :D (but maybe add a note on the Get saying that this may return values out of order, though?)
P.S. I know that this is just code being moved, but thought it'd be interesting to mention.
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Get
doesn't actually return values. It simply returns the channel and it is the caller's responsibility to read from that channel. So, in your example when1.3
returns, it does not mean that the underlying channel in the buffer implementation has been drained. It just means thatgoroutine-1
has a channel from which it can read the same two values that it pushed.There was a problem hiding this comment.
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I understand that, but to clarify, I meant
x := <-b.Get()
when I said "1.3. Get()":The question is: is
v2
guaranteed to be 2?I think
v2 = 3
,v3 = 2
is possible ifb.Put(3)
happens before the post-barrierb.Load()
, which breaks ordering, no? Because2
was inserted before3
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You are right with your last statement, but I don't think that breaks ordering. This buffer has the same behavior as a buffered channel. In your example, if you used a vanilla channel with a buffer of 3, you can still run into the same issue that you have mentioned here.
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True. I assumed that since
Put
is non-blocking (unlike a vanilla channel), a user might expect the result to be in the order thatPut
s returned (like a vanilla channel). That is, in a vanilla channel of capacity 2, ifch <- x
completes beforech <- y
, then receiving from the channel guarantees thatx
will be returned beforey
. This buffer, however, does not guarantee it, correct?There was a problem hiding this comment.
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Note that I impose no restriction on which channel send completes first (that's entirely random); just that if
x
completes first, thenx
will be returned first when receiving from the channel.There was a problem hiding this comment.
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Coming back to your original question:
If you can guarantee that the assignment to
v2
happens before the assignment tov3
, then we can guarantee thatv2 will be equal to 2
.If you look at it from the point of view of reading from the channel returned from
Get
, it still maintains order. Now, if you use the returned channel in different goroutines, the order in which you see values will be the order in which the reads execute.Maybe I'm missing your point.