5 ms·

As long as it's a fairly consistently timed window (and I assume it is), it doesn't matter if it's just a few milliseconds. On a related note, I'm still surpri

As long as it's a fairly consistently timed window (and I assume it is), it doesn't matter if it's just a few milliseconds. On a related note, I'm still surprised to find that telephone operators haven't fixed the scenario "someone calls me and I just miss it, I call them back while they're trying to call me for the second time, and we both get a busy signal". As a naive outsider, seems like it wouldn't take much to just connect those two calls.

Sir_Substance 9y ago
>As a naive outsider, seems like it wouldn't take much to just connect those two calls. Who would you charge for the call?
- kaslai 9y ago
  All phones services I've used charge for total usage, regardless of the initiator. If someone calls me, I still get dinged for every minute I'm on the line, same as if I placed an outgoing call. (Granted, I've only used pay-as-you-go type services, so perhaps other services are different.)
  - josteink 9y ago
    In Europe all services I have seen has been of the kind where whoever makes the call pays, and nobody ever gets charged for receiving calls. Under such a system, this can become a issue.
    - Moru 9y ago
      I bet from the phone company they don't see the problem. Both were intending to place a call so they both pay :)
  - threeseed 9y ago
    Apparently only the US and Canada charge the phone owner for incoming calls. https://en.wikipedia.org/wiki/GSM_services#Incoming_calls https://en.wikipedia.org/wiki/GSM_services#Incoming_calls Europe, Asia, Australia etc all charge the caller.
    - kalleboo 9y ago
      This leads to awkward situations like when I called someone in the US and they were worried that they're going to get charged international rates for a call from abroad.
    - Sir_Substance 9y ago
      Wow, dick move. Getting charged for the incoming call is something I associate with getting ripped off by roaming call charges.
  - therealidiot 9y ago
    Many years ago, using Virgin Mobile (in the UK) I remember being charged for incoming calls, however these days on all UK networks (and I assume this is the case for most of Europe) the caller gets charged but the callee does not
- ForHackernews 9y ago
  Both of them, obviously. ;)
- wruza 9y ago
  The one with lower timestamp.
derefr 9y ago
Under that scheme, if I called my own line by dialing my own DID, I'd end up with a completed loopback call rather than a connection to my voicemail inbox.
- eru 9y ago
  Just have an exception for that case?
- PrimHelios 9y ago
  I don't know how phone systems work, but in any other programming scenario, a quick if..else would fix that.
  - derefr 9y ago
    To translate into Internet-engineering terms: assume that you might be having to call out through a proxy server (this is true when you e.g. have to "dial 9" to reach an outside line.) If your own phone is public-routable (i.e. your extension has its own DID, and that's what you're calling), then the routing system outside the device will have no idea that what you'll reach is, in fact, your own device. The origin of the call will appear to the POTS system as just "the PBX" (a NAT, basically); and the destination will appear to your PBX as just some anonymous public number owned by a phone company, that happens to proxy back to your own extension on the PBX when you request it. So neither the POTS system nor your PBX will be able to recognize what's going on. (I mean, you can manually program associated numbers into the PBX, but by doing so you're denormalizing DID mappings: they'll break as soon as someone's number changes on the POTS side.) Much better if dialing out on a line just takes an exclusionary lock on the line so that it can't be dialed into; then anything that forwards back to that line, at any level of indirection, will be told the line is busy and will step to the next part of the call-tree. (Note that this becomes important when you're trying to e.g. get a [random peer of your support-level] on the line for a consult in tech support, by "adding" to the call the same ring-group you're in. Exclusionary locking here is what ensures that you won't end up adding yourself to the call.
    - 9y ago

Sir_Substance 9y ago

>As a naive outsider, seems like it wouldn't take much to just connect those two calls. Who would you charge for the call?

kaslai 9y ago
All phones services I've used charge for total usage, regardless of the initiator. If someone calls me, I still get dinged for every minute I'm on the line, same as if I placed an outgoing call. (Granted, I've only used pay-as-you-go type services, so perhaps other services are different.)
- josteink 9y ago
  In Europe all services I have seen has been of the kind where whoever makes the call pays, and nobody ever gets charged for receiving calls. Under such a system, this can become a issue.
  - Moru 9y ago
    I bet from the phone company they don't see the problem. Both were intending to place a call so they both pay :)
- threeseed 9y ago
  Apparently only the US and Canada charge the phone owner for incoming calls. https://en.wikipedia.org/wiki/GSM_services#Incoming_calls https://en.wikipedia.org/wiki/GSM_services#Incoming_calls Europe, Asia, Australia etc all charge the caller.
  - kalleboo 9y ago
    This leads to awkward situations like when I called someone in the US and they were worried that they're going to get charged international rates for a call from abroad.
  - Sir_Substance 9y ago
    Wow, dick move. Getting charged for the incoming call is something I associate with getting ripped off by roaming call charges.
- therealidiot 9y ago
  Many years ago, using Virgin Mobile (in the UK) I remember being charged for incoming calls, however these days on all UK networks (and I assume this is the case for most of Europe) the caller gets charged but the callee does not
ForHackernews 9y ago
Both of them, obviously. ;)
wruza 9y ago
The one with lower timestamp.

derefr 9y ago

Under that scheme, if I called my own line by dialing my own DID, I'd end up with a completed loopback call rather than a connection to my voicemail inbox.

eru 9y ago
Just have an exception for that case?
PrimHelios 9y ago
I don't know how phone systems work, but in any other programming scenario, a quick if..else would fix that.
- derefr 9y ago
  To translate into Internet-engineering terms: assume that you might be having to call out through a proxy server (this is true when you e.g. have to "dial 9" to reach an outside line.) If your own phone is public-routable (i.e. your extension has its own DID, and that's what you're calling), then the routing system outside the device will have no idea that what you'll reach is, in fact, your own device. The origin of the call will appear to the POTS system as just "the PBX" (a NAT, basically); and the destination will appear to your PBX as just some anonymous public number owned by a phone company, that happens to proxy back to your own extension on the PBX when you request it. So neither the POTS system nor your PBX will be able to recognize what's going on. (I mean, you can manually program associated numbers into the PBX, but by doing so you're denormalizing DID mappings: they'll break as soon as someone's number changes on the POTS side.) Much better if dialing out on a line just takes an exclusionary lock on the line so that it can't be dialed into; then anything that forwards back to that line, at any level of indirection, will be told the line is busy and will step to the next part of the call-tree. (Note that this becomes important when you're trying to e.g. get a [random peer of your support-level] on the line for a consult in tech support, by "adding" to the call the same ring-group you're in. Exclusionary locking here is what ensures that you won't end up adding yourself to the call.
  - 9y ago