Out of curiosity do you do something like this in wireshark?
ip.dst_host=="duckduckgo.com" && tcp.flags.syn == 0
My wireshark fu is really weak...
tors 3 aug. 2017 kl 11:35 skrev :
> Works, thanks a lot!!!
> Also found this related discussion:
> https://github.com/golang/go/issues/13397
>
> BR
On Thursday, August 3, 2017 at 11:07:50 AM UTC+2, pala.d...@gmail.com wrote:
>
> Example code with DuckDuckGo as default target:
>
> https://play.golang.org/p/_iI5-MDJ5t
>
This looks like a thundering herd race on connection setup. So there is no
connection to reuse as none of them has been setup
What is your target service you are calling? Is it available online?
tors 3 aug. 2017 kl 09:46 skrev :
>
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to golang
A goroutine making each request, passing in the http.Client should work.
Something like http://blog.narenarya.in/concurrent-http-in-go.html
On Wed, 2 Aug 2017 at 18:58 wrote:
> Parallel in the sense of HTTP/2 multiplexing, i.e. sending the next
> request without having to wait for the reply of
Parallel in the sense of HTTP/2 multiplexing, i.e. sending the next request
without having to wait for the reply of the previous one.
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from i
Parallel, on the same connection? How?
Which byte shall be sent next?
TCP is essentially sequential, sorry.
--
You received this message because you are subscribed to the Google Groups
"golang-nuts" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to golang
On Thursday, 13 July 2017 15:56:27 UTC+1, pala.d...@gmail.com wrote:
>
> Hello, I'm trying to understand how to write an HTTP client using the
> standard HTTP package and which takes advantage of HTTP/2 multiplexing
> (i.e. send many requests "in parallel" using the same TCP connection).
>
> On
