What "Ian Denhardt" says is quite reasonnable below.
>
>
>
>
>
> *The error you're seeing isn't an "out of bounds" error, rather, on
> the last send the channel is already full so the sending goroutine waits
> until there is space to insert the new item. Since there is nothing that
> will ever t
This looks quite interesting for the types of simple apps I'm building
right now. However, I have discovered a couple of bugs. I don't see any way
to report them on bitbucket.
Are you taking bug reports, and if so, what is the best way to communicate
them?
Are you taking outside pull requests
On 29. jan. 2019, at 2:00 f.h., 伊藤和也 wrote:
> In general computing, a deadlock is a situation where two different programs
> or processes depend on one another for completion, either because both are
> using the same resources or because of erroneous cues or other
> problems.https://www.techo
Quoting 伊藤和也 (2019-01-29 05:00:32)
>I know the general meaning of a deadlock, but I don't know the meaning
>of a deadlock in golang. For example, I send 4 values to a buffered
>channel whose maxmum size is 3 and a deadlock occurs. I think this is
>just "values are out of bounds" li
On Mon, Jan 28, 2019 at 11:53 PM Aakash das wrote:
>
> +golang-nuts.
>
> On Tue, Jan 29, 2019 at 1:20 PM Aakash das wrote:
>>
>> Thanks for the reply, Ian. I have an additional question. Given a Go binary
>> package file in '.a' format, would it be possible to extract dependencies
>> for the sa
On Tue, Jan 29, 2019 at 12:55 AM 伊藤和也 wrote:
>
> I know the general meaning of a deadlock, but I don't know the meaning of a
> deadlock in golang. For example, I send 4 values to a buffered channel whose
> maxmum size is 3 and a deadlock occurs. I think this is just "values are out
> of bounds"
If the server doesn't do an writes or is stuck in a read you can see
this behavior.
Enabling keepalive on the connection should fix this.
On 29/01/2019 10:57, mailme...@gmail.com wrote:
I wrote a both server and client side with golang, the server side
runs on centos7, and I built client side
I wrote a both server and client side with golang, the server side runs on
centos7, and I built client side as aar package runs at android, but I
noticed If I close android network, the server side websocket.read not get
a EOF error and it still keep hang.
What is the reason? how can I solve it
On Tue, Jan 29, 2019 at 10:59 AM 伊藤和也 wrote:
A deadlock is the same thing in any environment/any programming language:
Waiting for something to happen that will provably not happen.
So yes, deadlock in Go is the same as deadlock anywhere else.
PS: Please do not paste colorized code. Particularl
I quoted 3 explanation about a deadlock from 3 websites. The idea of the
deadlock in golang and the explanations are the same or similar or
different?
A deadlock is a situation in which two computer programs sharing the same
resource are effectively preventing each other from accessing the reso
I quoted 3 explanation about a deadlock from 3 websites. The ideo of the
deadlock in golang and the explanations are the same or similar or
different?
A deadlock is a situation in which two computer programs sharing the same
resource are effectively preventing each other from accessing the reso
Yes, exactly. It's a deadlock in the general sense, but this case is
particularly easy to detect: if all of the goroutines in your program
are blocked on some channel operation, then the runtime's scheduler
notices it has nothing it can do and reports that the system has
deadlocked.
-Ian
Quoting
It's because the goroutine executing the main function blocks until the last
channel send completes. But since you don't have another goroutine receiving
from the channel the program cannot continue and remains in a blocked state.
This is the cause of the deadlock.
On Tue, 29 Jan 2019, at 8:55
Hi, maybe it's because no goroutine read the channel, the four value
cause the deadlock because for write that value there is no space in the
channel, sorry by my English, I'm new in go too...
El mar., 29 ene. 2019 9:55, 伊藤和也 escribió:
> I know the general meaning of a deadlock, but I don't k
I know the general meaning of a deadlock, but I don't know the meaning of a
deadlock in golang. For example, I send 4 values to a buffered channel
whose maxmum size is 3 and a deadlock occurs. I think this is just "values
are out of bounds" like array. What does a deaklock mean in golang?
func
15 matches
Mail list logo
