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trying to understand golang chan cause crash or do something else
Adding a func never called improves behavior?Is blocking on a channel send a bad synchronization paradigm and whywhat would cause code blocking in go?What's the difference between c:=make(chan int) and c:=make(chan int,1)?Convert chan to non chan in golangGolang: Why does increasing the size of a buffered channel eliminate output from my goroutines?Golang - Range over chan map by keyWhy can I not use func literal as custom func type despite matching parameters in golang?Multiple go routines consuming from a channel causing loss of datareturn early from range of a channel
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
The following is a example from https://golang.org/ref/mem:
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.)
I understand that It might print the empty string, but not for crash or do something else, when will crash happen? And when will it do something else ?
go memory concurrency
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
add a comment |
The following is a example from https://golang.org/ref/mem:
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.)
I understand that It might print the empty string, but not for crash or do something else, when will crash happen? And when will it do something else ?
go memory concurrency
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
7
The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26
add a comment |
The following is a example from https://golang.org/ref/mem:
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.)
I understand that It might print the empty string, but not for crash or do something else, when will crash happen? And when will it do something else ?
go memory concurrency
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
The following is a example from https://golang.org/ref/mem:
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.)
I understand that It might print the empty string, but not for crash or do something else, when will crash happen? And when will it do something else ?
go memory concurrency
go memory concurrency
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
asked Mar 27 at 8:23
Geln YangGeln Yang
5392 gold badges15 silver badges35 bronze badges
5392 gold badges15 silver badges35 bronze badges
7
The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26
add a comment |
7
The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26
7
7
The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26
The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26
add a comment |
1 Answer
1
active
oldest
votes
A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
add a comment |
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A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
add a comment |
A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
add a comment |
A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
A string in Go is a read only slice of bytes. Slice consists of length an pointer. Let's assume that we first set length to a large value and then change the pointer. The other go routine may first read new length and old pointer. Then it tries to read over the end of the previous string. It either read some garbage or is stopped by operating system and crashes.
The order of operations does not matter really, if you set pointer firsts it may point to memory area too short for the current length.
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
edited Mar 27 at 9:48
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
answered Mar 27 at 8:36
Grzegorz ŻurGrzegorz Żur
33.2k13 gold badges85 silver badges82 bronze badges
33.2k13 gold badges85 silver badges82 bronze badges
add a comment |
add a comment |
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The behavior is undefined. Don't try to find logic in it. Read: Benign Data Races: What Could Possibly Go Wrong? And if not enogh: Golang data races to break memory safety
– icza
Mar 27 at 8:26