The Go Memory Model
Version of May 31, 2014
Introduction
The Go memory model specifies the conditions under which reads of a variable in one goroutine can be guaranteed to observe values produced by writes to the same variable in a different goroutine.
Advice
Programs that modify data being simultaneously accessed by multiple goroutines must serialize such access.
To serialize access, protect the data with channel operations or other synchronization primitives
such as those in the sync
and sync/atomic
packages.
If you must read the rest of this document to understand the behavior of your program, you are being too clever.
Don't be clever.
Happens Before
Within a single goroutine, reads and writes must behave
as if they executed in the order specified by the program.
That is, compilers and processors may reorder the reads and writes
executed within a single goroutine only when the reordering
does not change the behavior within that goroutine
as defined by the language specification.
Because of this reordering, the execution order observed
by one goroutine may differ from the order perceived
by another. For example, if one goroutine
executes a = 1; b = 2;
, another might observe
the updated value of b
before the updated value of a
.
To specify the requirements of reads and writes, we define happens before, a partial order on the execution of memory operations in a Go program. If event e1 happens before event e2, then we say that e2 happens after e1. Also, if e1 does not happen before e2 and does not happen after e2, then we say that e1 and e2 happen concurrently.
Within a single goroutine, the happens-before order is the order expressed by the program.
A read r of a variable v
is allowed to observe a write w to v
if both of the following hold:
- r does not happen before w.
- There is no other write w' to
v
that happens after w but before r.
To guarantee that a read r of a variable v
observes a
particular write w to v
, ensure that w is the only
write r is allowed to observe.
That is, r is guaranteed to observe w if both of the following hold:
- w happens before r.
- Any other write to the shared variable
v
either happens before w or after r.
This pair of conditions is stronger than the first pair; it requires that there are no other writes happening concurrently with w or r.
Within a single goroutine,
there is no concurrency, so the two definitions are equivalent:
a read r observes the value written by the most recent write w to v
.
When multiple goroutines access a shared variable v
,
they must use synchronization events to establish
happens-before conditions that ensure reads observe the
desired writes.
The initialization of variable v
with the zero value
for v
's type behaves as a write in the memory model.
Reads and writes of values larger than a single machine word behave as multiple machine-word-sized operations in an unspecified order.
Synchronization
Initialization
Program initialization runs in a single goroutine, but that goroutine may create other goroutines, which run concurrently.
If a package p
imports package q
, the completion of
q
's init
functions happens before the start of any of p
's.
The start of the function main.main
happens after
all init
functions have finished.
Goroutine creation
The go
statement that starts a new goroutine
happens before the goroutine's execution begins.
For example, in this program:
var a string func f() { print(a) } func hello() { a = "hello, world" go f() }
calling hello
will print "hello, world"
at some point in the future (perhaps after hello
has returned).
Goroutine destruction
The exit of a goroutine is not guaranteed to happen before any event in the program. For example, in this program:
var a string func hello() { go func() { a = "hello" }() print(a) }
the assignment to a
is not followed by
any synchronization event, so it is not guaranteed to be
observed by any other goroutine.
In fact, an aggressive compiler might delete the entire go
statement.
If the effects of a goroutine must be observed by another goroutine, use a synchronization mechanism such as a lock or channel communication to establish a relative ordering.
Channel communication
Channel communication is the main method of synchronization between goroutines. Each send on a particular channel is matched to a corresponding receive from that channel, usually in a different goroutine.
A send on a channel happens before the corresponding receive from that channel completes.
This program:
var c = make(chan int, 10) var a string func f() { a = "hello, world" c <- 0 } func main() { go f() <-c print(a) }
is guaranteed to print "hello, world"
. The write to a
happens before the send on c
, which happens before
the corresponding receive on c
completes, which happens before
the print
.
The closing of a channel happens before a receive that returns a zero value because the channel is closed.
In the previous example, replacing
c <- 0
with close(c)
yields a program with the same guaranteed behavior.
A receive from an unbuffered channel happens before the send on that channel completes.
This program (as above, but with the send and receive statements swapped and using an unbuffered channel):
var c = make(chan int) var a string func f() { a = "hello, world" <-c } func main() { go f() c <- 0 print(a) }
is also guaranteed to print "hello, world"
. The write to a
happens before the receive on c
, which happens before
the corresponding send on c
completes, which happens
before the print
.
If the channel were buffered (e.g., c = make(chan int, 1)
)
then the program would not be guaranteed to print
"hello, world"
. (It might print the empty string,
crash, or do something else.)
The kth receive on a channel with capacity C happens before the k+Cth send from that channel completes.
This rule generalizes the previous rule to buffered channels. It allows a counting semaphore to be modeled by a buffered channel: the number of items in the channel corresponds to the number of active uses, the capacity of the channel corresponds to the maximum number of simultaneous uses, sending an item acquires the semaphore, and receiving an item releases the semaphore. This is a common idiom for limiting concurrency.
This program starts a goroutine for every entry in the work list, but the
goroutines coordinate using the limit
channel to ensure
that at most three are running work functions at a time.
var limit = make(chan int, 3) func main() { for _, w := range work { go func(w func()) { limit <- 1 w() <-limit }(w) } select{} }
Locks
The sync
package implements two lock data types,
sync.Mutex
and sync.RWMutex
.
For any sync.Mutex
or sync.RWMutex
variable l
and n < m,
call n of l.Unlock()
happens before call m of l.Lock()
returns.
This program:
var l sync.Mutex var a string func f() { a = "hello, world" l.Unlock() } func main() { l.Lock() go f() l.Lock() print(a) }
is guaranteed to print "hello, world"
.
The first call to l.Unlock()
(in f
) happens
before the second call to l.Lock()
(in main
) returns,
which happens before the print
.
For any call to l.RLock
on a sync.RWMutex
variable l
,
there is an n such that the l.RLock
happens (returns) after call n to
l.Unlock
and the matching l.RUnlock
happens
before call n+1 to l.Lock
.
Once
The sync
package provides a safe mechanism for
initialization in the presence of multiple goroutines
through the use of the Once
type.
Multiple threads can execute once.Do(f)
for a particular f
,
but only one will run f()
, and the other calls block
until f()
has returned.
A single call of f()
from once.Do(f)
happens (returns) before any call of once.Do(f)
returns.
In this program:
var a string var once sync.Once func setup() { a = "hello, world" } func doprint() { once.Do(setup) print(a) } func twoprint() { go doprint() go doprint() }
calling twoprint
will call setup
exactly
once.
The setup
function will complete before either call
of print
.
The result will be that "hello, world"
will be printed
twice.
Incorrect synchronization
Note that a read r may observe the value written by a write w that happens concurrently with r. Even if this occurs, it does not imply that reads happening after r will observe writes that happened before w.
In this program:
var a, b int func f() { a = 1 b = 2 } func g() { print(b) print(a) } func main() { go f() g() }
it can happen that g
prints 2
and then 0
.
This fact invalidates a few common idioms.
Double-checked locking is an attempt to avoid the overhead of synchronization.
For example, the twoprint
program might be
incorrectly written as:
var a string var done bool func setup() { a = "hello, world" done = true } func doprint() { if !done { once.Do(setup) } print(a) } func twoprint() { go doprint() go doprint() }
but there is no guarantee that, in doprint
, observing the write to done
implies observing the write to a
. This
version can (incorrectly) print an empty string
instead of "hello, world"
.
Another incorrect idiom is busy waiting for a value, as in:
var a string var done bool func setup() { a = "hello, world" done = true } func main() { go setup() for !done { } print(a) }
As before, there is no guarantee that, in main
,
observing the write to done
implies observing the write to a
, so this program could
print an empty string too.
Worse, there is no guarantee that the write to done
will ever
be observed by main
, since there are no synchronization
events between the two threads. The loop in main
is not
guaranteed to finish.
There are subtler variants on this theme, such as this program.
type T struct { msg string } var g *T func setup() { t := new(T) t.msg = "hello, world" g = t } func main() { go setup() for g == nil { } print(g.msg) }
Even if main
observes g != nil
and exits its loop,
there is no guarantee that it will observe the initialized
value for g.msg
.
In all these examples, the solution is the same: use explicit synchronization.