Wikipedia:Reference desk/Archives/Computing/2007 July 29

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July 29[edit]

Where can I find linux code?[edit]

Linus Torvalds developed the Linux operating system to where I can find of that? (I heard that he has development of the Linux operating system is a heap of code to enable others to develop) Incidentally, If only a heap of code it, then how they develop into a type of operating system?

Read the article on Linux (and the relative articles). They contain the information you want. --racergr 06:44, 29 July 2007 (UTC)[reply]
If with your second question you mean to ask how little snippets of code can develop into something as big as an OS, then the answer is modular programming. Make sure that that each snippet is completely separate and clear about its task, ie what are the precise input and output. Then, you can make bigger programs using those snippets, which also abide by the aforementioned rules. So modules are built out of modules etc, until you've made something very big out of lots of little things. DirkvdM 06:58, 29 July 2007 (UTC)[reply]
Linux is NOT an operating system! If you're looking for the source code for the Linux kernel, go to kernel.org. --wj32 talk | contribs 09:46, 29 July 2007 (UTC)[reply]
Yes, which like the first post said, is answered by some reading of our articles on linux. Either way, anyone that doesn't recognize linux is technically the kernel probably won't get any use out of it --Lie! 11:53, 29 July 2007 (UTC)[reply]
If you want to dive into code, you can browse the linux cross referencer --Ancheta Wis 22:37, 29 July 2007 (UTC)[reply]

Difference between 2 and 2&3 in a shift operation.[edit]

I have come across this code when looking on the XXTEA reference code:

e = sum >> 2&3;

As far as I know it essentialy means:

e = sum >> 2;

So I am wondering why did the developers of XXTEA choose to use an extra AND instruction? What did I miss here? --racergr 06:49, 29 July 2007 (UTC)[reply]

It's probably (sum >> 2) & 3, where there's a bitmask involved.
Indeed, because >> has higher precedence than &. This is a great example of why brackets should be used when there's any room for ambiguity; even writing e = sum>>2 &3 rather than e = sum >> 2&3 would have been better. -- Finlay McWalter | Talk 13:37, 29 July 2007 (UTC)[reply]

System Memory on Mac OS X[edit]

When I use Activity Monitor to look at the system memory, it lists the memory used in four groups: Wired, active, inactive, free. I can guess what free means, but what do the other three groups refer to? 69.123.113.89 11:14, 29 July 2007 (UTC)[reply]

Try a few of these links. Weregerbil 15:06, 29 July 2007 (UTC)[reply]
Thanks. 69.123.113.89 15:58, 29 July 2007 (UTC)[reply]

Router IP assignment[edit]

Just wondering, if you have 2 routers connected to each other to form a local network, what happens with address assignment? How is it decided which one gets to assign local IP addresses? My example is: I have a router for ADSL2, but am getting a new one to do wireless networking (but it doesn't do ADSL decoding) so I have to keep them both. So, how is it decided which of the routers assigns addresses? - EstoyAquí(tce) 11:53, 29 July 2007 (UTC)[reply]

Two-part answer:
  1. Address assignment isn't a necessary function of a router. Routers do routing; address assignment is done by a DHCP server. (Of course, there's nothing that says that a single box can't act as both a router and a DHCP server, and indeed many home-based "routers" do both, and more.)
  2. If your router is a true router, then by definition you have two separate network segments, which it's the router's job to route packets between. If you have two routers, it sounds like you have at least three network segments. Furthermore, DHCP (or indeed any addressing scheme) must be applied to individual network segments individually; indeed one way to define a network segment is "a subset of the larger internetwork having its own unique subset of addresses and its own addressing scheme". So DHCP packets, by design, travel only on individual network segments; they are not passed between segments by routers. So assuming that your two DHCP servers are assigning addresses in two different of your network segments, all should be well.
Steve Summit (talk) 13:23, 29 July 2007 (UTC)[reply]
Ok thanks. So, are you saying that DHCP is what assigns the addresses? As far as I know they both do DHCP. So does that mean, to disable DHCP on one would allow the other to assign all addresses? - EstoyAquí(tce) 13:27, 29 July 2007 (UTC)[reply]
Yes. Well, probably. Well, maybe. A definitive answer would require knowing the details of your "network topology": how these two routers are connected to each other and to your "outbound" Internet connection, and where and how you'll be connecting the computers that need address assignment. —Steve Summit (talk) 13:30, 29 July 2007 (UTC)[reply]

mother of computer[edit]

we know that charles babbage is the father of computer then who is the mother of computer?

Ada Lovelace -- Finlay McWalter | Talk 13:33, 29 July 2007 (UTC)[reply]
Or maybe Grace Murray Hopper. -- Finlay McWalter | Talk 13:39, 29 July 2007 (UTC)[reply]
I usually say that Babbage and Lovelace are the grandparents and Alan Turing and Grace Hopper are the parents. Or something... --Oskar 19:50, 29 July 2007 (UTC)[reply]
Would that make John Von Neumann an uncle? Or can we have more than two parents? --24.147.87.78 20:53, 29 July 2007 (UTC)[reply]
He's like the cousin you have weird feelings for that suppress because you know it can never happen --Oskar 22:37, 29 July 2007 (UTC)[reply]
Actually, now that I think of it, Grace Hopper and John von Neumann are the parents of computing, because then Alan Turing can be the uncle everyone knows is gay but loves anyway. --Oskar 22:39, 29 July 2007 (UTC)[reply]

Cannot apt-get update in Debian[edit]

Some problem I'm having, I just installed Debian on an old computer using a 1st generation Pentium 4 (1.5 GHz) and I'm trying to apt-get update it, like some of the guides on the internet I've found said. The main problem is, having to deal with Debian defaulting to some IPv6 protocol, I keep getting this:

Could not connect to ftp.tw.debian.org:80 (1.0.0.0), connection timed out

I ever had this problem on Windows XP; I installed ALL the Networking Services tool on the old computer and then I was TOTALLY LOCKED OUT from the internet!!! Internet Explorer's diagnostic test can ping through while Firefox just shows it's timeout error. Everything keeps resolving to the 1.0.0.0 IP. Checking through ipconfig, there's some strange adapters listed: Teredo Automatic Tunneling Device if I can remember the name and another one. The IP's are some uncommon hex notation, ff28:5a9c:2abc:9256 as an example. I finally did a System Restore operation and I can access the internet like before.

I've searched the internet for this problem and there are many cases listed, but since I cannot find mine with my case, I just decided that I should just post here. I'm using an Aztech DSL600ER router, Linux OS is Debian, and I can access Wikipedia after I changed the network.dns.disableIPv6 value to FALSE. Other than that, to reiterate the topic name, I cannot apt-get update through the terminal as root user. Is there any way to resolve this 1.0.0.0 IP problem? --Bruin_rrss23 (talk) 14:26, 29 July 2007 (UTC)[reply]

I'm not really familiar with this, but I'd look up instructions for disabling IPv6 for your distro, that sounds like the problem to me. That hex notation is an IPv6 address, which doesn't really work yet in most of the world. --Lie! 14:30, 29 July 2007 (UTC)[reply]
Looks like it's not a IPv6 problem, but a buggy home router (see [1] for instance). Try setting the DNS servers manually on /etc/resolv.conf; according to that bug report, simply disabling IPv6 is not enough to make it work. The best option would be to either fix the router or change to one which doesn't have the bug. --cesarb 22:09, 29 July 2007 (UTC)[reply]
Another option, by the way, would be to (if the option is available on your home router) change it to be only a "bridge", and run the PPPoE (if it's ADSL) stack directly on your host. That way, you'd bypass almost all the potentially buggy software on your home router (and also gain a lot of extra control). Be sure to configure a firewall before doing it, since it usually would make your computer directly visible to the whole Internet. --cesarb 22:26, 29 July 2007 (UTC)[reply]
Check your default route.
Thanks for your replies, but since I'm using BitTorrent on another computer connected to the same router, I just tried all the software solutions without any success. It just shows the normal IP, then reverts back to 1.0.0.0. It manages to get through but I get cannot connect to host. I just kept modding the files as the instructions asked me to so I can get out of the 1.0.0.0 situation but Debian cannot get into the root account, so I just had to do a complete reinstall. Maybe the developers at Debian.org should make the next release have IPv6 DISABLED by default. Thanks for your replies anyway. I'll keep trying to solve this annoying issue. Any help would be appreciated. --Bruin_rrss23 (talk) 09:27, 31 July 2007 (UTC)[reply]
Having IPv6 disabled by default wouldn't help in this case; the problem is with the buggy DNS proxy on the router, which runs over IPv4. There's another thing you could try; install BIND on the machine itself (apt-get install bind9), and point its /etc/resolv.conf at 127.0.0.1 (nameserver 127.0.0.1 and remove all the other nameserver lines; note that it might be automatically recreated with the original contents via DHCP). As long as the router doesn't try to be even more annoying by doing transparent proxying of DNS requests, that should be enough to give you a clean working DNS setup. --cesarb 23:46, 31 July 2007 (UTC)[reply]

Office 2007 Save As glitch[edit]

Hello. Sometimes when I am working on a Word document on Office 2007, I save my file. It is not a read-only before I enter it. The Save As window appears. When I try to replace my file with the same name, Office says the file is a read-only and must be saved under a different name. How can I fix this problem? Thanks in advance. --Mayfare 15:23, 29 July 2007 (UTC)[reply]

ActionScript 3 question[edit]

How can I get the color of a given shape object in Flash using ActionScript 3? That is — I have a timeline with nothing on it but a single shape which shape tweens between two colors (say, red and blue) over the course of 25 frames or so. How can I, say, get the color of the shape at an arbitrary frame setting? The shape is not an object — it is just a vector shape.

My feeling is that I could somehow just grab a reference to the shape by playing with this.stage.getChildAt(0) but I don't think that does what I think it does. Any thoughts?

The reason I'm trying to do this is that I'd like to be able to assign a color to a different object based on the colors generated in this other movieclip via tweening, etc. There are a number of reasons why I can't just tween the original object itself (the main is that it is already shape tweening and that shape tweening must be independent of the color change).

Does this make any sense? --66.30.5.140 20:00, 29 July 2007 (UTC)[reply]

OK — I've figured out how to point to the objects in question (by using this.getChildAt(0)), and they are MorphShapes while tweening. Now I see that interactions with MorphShapes are pretty limited — but is there any way to get its fill color? --66.30.5.140 20:26, 29 July 2007 (UTC)[reply]
OK — I think, after much searching, that the answer is definitely no, so I found a different approach to my issue (not as elegant as I would have wanted, but it works) that avoids this method entirely. --24.147.86.187 00:53, 30 July 2007 (UTC)[reply]

DirectX 10[edit]

Does anybody know if there's any chance of DirectX 10 being offered for XP? Vista hasn't exactly set the world on fire so I can't help but think that keeping it as a Vista exclusive until 7's release will only hurt DirectX 10 as many developers may not feel the need to include features that the majority of customers will never see. --Kiltman67 20:14, 29 July 2007 (UTC)[reply]

Seems unlikely to me, particularly given that DX10 exclusive games are still a very, very long way away given that DX10 cards are still very new Nil Einne 23:46, 29 July 2007 (UTC)[reply]
Microsoft develop DX10. Microsoft want to sell Vista. Put those 2 together and it's obvious that it's unlikely to happen. Capuchin 08:04, 30 July 2007 (UTC)[reply]
I believe there's a company that's selling a product that will allow you to run Halo 2 on XP, although I don't think that's a DX10 game anyway. It could go either way for MS, having a tight monopoly has worked pretty well in the past for them. Personally, I'm rooting for Mt. Evans --Lie! 08:11, 30 July 2007 (UTC)[reply]

DirectX 10 simply can not work in XP, for reasons related to the new driver framework in Vista, which would be entirely self-defeating to implement in XP.The reason that Shadowrun and Halo 2 can work in Single-player in XP is that they aren't DirectX 10 games (If they were, you wouldn't be able to play them on currently available hardware). For a complete, in-depth explanation of this, browse around staring from here: [2] --Alph Tech STUART 17:08, 4 August 2007 (UTC)[reply]

Picking A GPS[edit]

Hello everyone!

I am going to purchase an automotive GPS system, and I've narrowed it down to four units, all Garmin: StreetPilot c550, Nuvi 350, Nuvi 650, and Nuvi 660. I couldn't find any difference between the Nuvi 650 and the 660, except the Bluetooth compatibility and the price. How about the traffic situation? I know that some, traffic is included, just with a monthly fee. Others require costly equipment. I didn't consider any of the Magellans after hearing many horror stories, and the Tom Tom 910 was quickly thrown out. I am looking for basic functionality with some perks. I also couldn't find out which ones had which basemaps. Which one is best suited to me? Thanks!!! Location: Long Island, NY

MAP91 21:00, 29 July 2007 (UTC)[reply]