what is the difference between RLock() and Lock() in Golang?
up vote
-3
down vote
favorite
what is the difference between RLock() and Lock() in Golang and how they can be used efficiently when we use mutex Lock ?
go mutex
add a comment |
up vote
-3
down vote
favorite
what is the difference between RLock() and Lock() in Golang and how they can be used efficiently when we use mutex Lock ?
go mutex
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
what is the difference between RLock() and Lock() in Golang and how they can be used efficiently when we use mutex Lock ?
go mutex
what is the difference between RLock() and Lock() in Golang and how they can be used efficiently when we use mutex Lock ?
go mutex
go mutex
asked Nov 22 at 9:36
Sharath BJ
49144
49144
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Lock(): only one go routine read/write at a time by acquiring the lock.
RLock(): multiple go routine can read(not write) at a time by acquiring the lock.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
a := 0
lock := sync.RWMutex{}
for i := 1; i < 10; i++ {
go func(i int) {
lock.Lock()
fmt.Printf("Lock: from go routine %d: a = %dn",i, a)
time.Sleep(time.Second)
lock.Unlock()
}(i)
}
b := 0
for i := 11; i < 20; i++ {
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b)
time.Sleep(time.Second)
lock.RUnlock()
}(i)
}
<-time.After(time.Second*10)
}
1) When a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens?
- To acquire a Lock() for write it has to wait until RUnlock()
2) What happens when someone already acquired Lock() for map ,will other go-routine can still get RLock()
- if someone X already acquired Lock(), then other go-routine to get RLock() will have to wait until X release lock (Unlock())
3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
- Map is not thread safe. so "concurrent read/write of Map" can cause error.
See following example for more clarification:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
lock := sync.RWMutex{}
b := map[string]int{}
b["0"] = 0
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["0"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(1)
go func(i int) {
lock.Lock()
b["2"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["2"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(2)
<-time.After(time.Second*8)
fmt.Println("*************************************8")
go func(i int) {
lock.Lock()
b["3"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(3)
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(4)
<-time.After(time.Second*8)
}
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
add a comment |
up vote
0
down vote
A RWMutex is a reader/writer mutual exclusion lock. The lock can be held by an arbitrary number of readers or a single writer. The zero value for a RWMutex is an unlocked mutex.
A RWMutex must not be copied after first use.
If a goroutine holds a RWMutex for reading and another goroutine might call Lock, no goroutine should expect to be able to acquire a read lock until the initial read lock is released. In particular, this prohibits recursive read locking. This is to ensure that the lock eventually becomes available; a blocked Lock call excludes new readers from acquiring the lock.
A Mutex is a mutual exclusion lock. The zero value for a Mutex is an unlocked mutex.
The golang provide the channel is the best practice for concurrency control, so i think the efficiently way using sync.lock is not used it, use channel instead.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Lock(): only one go routine read/write at a time by acquiring the lock.
RLock(): multiple go routine can read(not write) at a time by acquiring the lock.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
a := 0
lock := sync.RWMutex{}
for i := 1; i < 10; i++ {
go func(i int) {
lock.Lock()
fmt.Printf("Lock: from go routine %d: a = %dn",i, a)
time.Sleep(time.Second)
lock.Unlock()
}(i)
}
b := 0
for i := 11; i < 20; i++ {
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b)
time.Sleep(time.Second)
lock.RUnlock()
}(i)
}
<-time.After(time.Second*10)
}
1) When a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens?
- To acquire a Lock() for write it has to wait until RUnlock()
2) What happens when someone already acquired Lock() for map ,will other go-routine can still get RLock()
- if someone X already acquired Lock(), then other go-routine to get RLock() will have to wait until X release lock (Unlock())
3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
- Map is not thread safe. so "concurrent read/write of Map" can cause error.
See following example for more clarification:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
lock := sync.RWMutex{}
b := map[string]int{}
b["0"] = 0
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["0"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(1)
go func(i int) {
lock.Lock()
b["2"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["2"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(2)
<-time.After(time.Second*8)
fmt.Println("*************************************8")
go func(i int) {
lock.Lock()
b["3"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(3)
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(4)
<-time.After(time.Second*8)
}
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
add a comment |
up vote
0
down vote
accepted
Lock(): only one go routine read/write at a time by acquiring the lock.
RLock(): multiple go routine can read(not write) at a time by acquiring the lock.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
a := 0
lock := sync.RWMutex{}
for i := 1; i < 10; i++ {
go func(i int) {
lock.Lock()
fmt.Printf("Lock: from go routine %d: a = %dn",i, a)
time.Sleep(time.Second)
lock.Unlock()
}(i)
}
b := 0
for i := 11; i < 20; i++ {
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b)
time.Sleep(time.Second)
lock.RUnlock()
}(i)
}
<-time.After(time.Second*10)
}
1) When a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens?
- To acquire a Lock() for write it has to wait until RUnlock()
2) What happens when someone already acquired Lock() for map ,will other go-routine can still get RLock()
- if someone X already acquired Lock(), then other go-routine to get RLock() will have to wait until X release lock (Unlock())
3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
- Map is not thread safe. so "concurrent read/write of Map" can cause error.
See following example for more clarification:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
lock := sync.RWMutex{}
b := map[string]int{}
b["0"] = 0
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["0"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(1)
go func(i int) {
lock.Lock()
b["2"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["2"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(2)
<-time.After(time.Second*8)
fmt.Println("*************************************8")
go func(i int) {
lock.Lock()
b["3"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(3)
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(4)
<-time.After(time.Second*8)
}
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Lock(): only one go routine read/write at a time by acquiring the lock.
RLock(): multiple go routine can read(not write) at a time by acquiring the lock.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
a := 0
lock := sync.RWMutex{}
for i := 1; i < 10; i++ {
go func(i int) {
lock.Lock()
fmt.Printf("Lock: from go routine %d: a = %dn",i, a)
time.Sleep(time.Second)
lock.Unlock()
}(i)
}
b := 0
for i := 11; i < 20; i++ {
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b)
time.Sleep(time.Second)
lock.RUnlock()
}(i)
}
<-time.After(time.Second*10)
}
1) When a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens?
- To acquire a Lock() for write it has to wait until RUnlock()
2) What happens when someone already acquired Lock() for map ,will other go-routine can still get RLock()
- if someone X already acquired Lock(), then other go-routine to get RLock() will have to wait until X release lock (Unlock())
3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
- Map is not thread safe. so "concurrent read/write of Map" can cause error.
See following example for more clarification:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
lock := sync.RWMutex{}
b := map[string]int{}
b["0"] = 0
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["0"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(1)
go func(i int) {
lock.Lock()
b["2"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["2"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(2)
<-time.After(time.Second*8)
fmt.Println("*************************************8")
go func(i int) {
lock.Lock()
b["3"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(3)
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(4)
<-time.After(time.Second*8)
}
Lock(): only one go routine read/write at a time by acquiring the lock.
RLock(): multiple go routine can read(not write) at a time by acquiring the lock.
package main
import (
"fmt"
"sync"
"time"
)
func main() {
a := 0
lock := sync.RWMutex{}
for i := 1; i < 10; i++ {
go func(i int) {
lock.Lock()
fmt.Printf("Lock: from go routine %d: a = %dn",i, a)
time.Sleep(time.Second)
lock.Unlock()
}(i)
}
b := 0
for i := 11; i < 20; i++ {
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b)
time.Sleep(time.Second)
lock.RUnlock()
}(i)
}
<-time.After(time.Second*10)
}
1) When a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens?
- To acquire a Lock() for write it has to wait until RUnlock()
2) What happens when someone already acquired Lock() for map ,will other go-routine can still get RLock()
- if someone X already acquired Lock(), then other go-routine to get RLock() will have to wait until X release lock (Unlock())
3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
- Map is not thread safe. so "concurrent read/write of Map" can cause error.
See following example for more clarification:
package main
import (
"fmt"
"sync"
"time"
)
func main() {
lock := sync.RWMutex{}
b := map[string]int{}
b["0"] = 0
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["0"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(1)
go func(i int) {
lock.Lock()
b["2"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["2"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(2)
<-time.After(time.Second*8)
fmt.Println("*************************************8")
go func(i int) {
lock.Lock()
b["3"] = i
fmt.Printf("Lock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("Lock: from go routine %d: lock releasedn",i)
lock.Unlock()
}(3)
go func(i int) {
lock.RLock()
fmt.Printf("RLock: from go routine %d: b = %dn",i, b["3"])
time.Sleep(time.Second*3)
fmt.Printf("RLock: from go routine %d: lock releasedn",i)
lock.RUnlock()
}(4)
<-time.After(time.Second*8)
}
edited Nov 22 at 11:39
answered Nov 22 at 10:50
nightfury1204
1,40248
1,40248
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
add a comment |
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
thanks for the answer ,have some question 1)when a go-routine has already acquired a RLock(), can another go-routine acquire a Lock() for write or it has to wait until RUnlock() happens? 2) what happens when someone already acquired Lock() for map ,will other go-routine can still get RLock() 3) Assuming we are dealing with Maps here, is there any possibility of "concurrent read/write of Map" error can come?
– Sharath BJ
Nov 22 at 11:01
1
1
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
@SharathBJ updated my answer
– nightfury1204
Nov 22 at 11:39
add a comment |
up vote
0
down vote
A RWMutex is a reader/writer mutual exclusion lock. The lock can be held by an arbitrary number of readers or a single writer. The zero value for a RWMutex is an unlocked mutex.
A RWMutex must not be copied after first use.
If a goroutine holds a RWMutex for reading and another goroutine might call Lock, no goroutine should expect to be able to acquire a read lock until the initial read lock is released. In particular, this prohibits recursive read locking. This is to ensure that the lock eventually becomes available; a blocked Lock call excludes new readers from acquiring the lock.
A Mutex is a mutual exclusion lock. The zero value for a Mutex is an unlocked mutex.
The golang provide the channel is the best practice for concurrency control, so i think the efficiently way using sync.lock is not used it, use channel instead.
add a comment |
up vote
0
down vote
A RWMutex is a reader/writer mutual exclusion lock. The lock can be held by an arbitrary number of readers or a single writer. The zero value for a RWMutex is an unlocked mutex.
A RWMutex must not be copied after first use.
If a goroutine holds a RWMutex for reading and another goroutine might call Lock, no goroutine should expect to be able to acquire a read lock until the initial read lock is released. In particular, this prohibits recursive read locking. This is to ensure that the lock eventually becomes available; a blocked Lock call excludes new readers from acquiring the lock.
A Mutex is a mutual exclusion lock. The zero value for a Mutex is an unlocked mutex.
The golang provide the channel is the best practice for concurrency control, so i think the efficiently way using sync.lock is not used it, use channel instead.
add a comment |
up vote
0
down vote
up vote
0
down vote
A RWMutex is a reader/writer mutual exclusion lock. The lock can be held by an arbitrary number of readers or a single writer. The zero value for a RWMutex is an unlocked mutex.
A RWMutex must not be copied after first use.
If a goroutine holds a RWMutex for reading and another goroutine might call Lock, no goroutine should expect to be able to acquire a read lock until the initial read lock is released. In particular, this prohibits recursive read locking. This is to ensure that the lock eventually becomes available; a blocked Lock call excludes new readers from acquiring the lock.
A Mutex is a mutual exclusion lock. The zero value for a Mutex is an unlocked mutex.
The golang provide the channel is the best practice for concurrency control, so i think the efficiently way using sync.lock is not used it, use channel instead.
A RWMutex is a reader/writer mutual exclusion lock. The lock can be held by an arbitrary number of readers or a single writer. The zero value for a RWMutex is an unlocked mutex.
A RWMutex must not be copied after first use.
If a goroutine holds a RWMutex for reading and another goroutine might call Lock, no goroutine should expect to be able to acquire a read lock until the initial read lock is released. In particular, this prohibits recursive read locking. This is to ensure that the lock eventually becomes available; a blocked Lock call excludes new readers from acquiring the lock.
A Mutex is a mutual exclusion lock. The zero value for a Mutex is an unlocked mutex.
The golang provide the channel is the best practice for concurrency control, so i think the efficiently way using sync.lock is not used it, use channel instead.
answered Nov 22 at 9:50
L.Hao
71
71
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53427824%2fwhat-is-the-difference-between-rlock-and-lock-in-golang%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown