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.
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:
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:
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.
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.
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).
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 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 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.
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 causes "hello, world" to be printed twice.
The first call to doprint runs setup once.
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.