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  • Does Bing support anything like Google's First Click Free program?

    - by Dan Fabulich
    Google has a program for webmasters called First Click Free. To implement First Click Free, you need to allow all users who find a document on your site via Google search to see the full text of that document, even if they have not registered or subscribed to see that content. The user's first click to your content area is free. However, once that user clicks a link on the original page, you can require them to sign in or register to read further. The user must be able to see the full content of a multi-page article. You can allow this by displaying all content on a single page to both Googlebot and users. Alternatively, you can use cookies to make sure that a user can visit each page of a multi-page article before being asked for registration or payment. Does Bing support anything like this?

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  • Can i use aac in an commercial app for free?

    - by Jason123
    I was wondering if i can use the aac codec in my commercial app for free (through lgpl ffmpeg). It says on the wiki: No licenses or payments are required to be able to stream or distribute content in AAC format.[36] This reason alone makes AAC a much more attractive format to distribute content than MP3, particularly for streaming content (such as Internet radio). However, a patent license is required for all manufacturers or developers of AAC codecs. For this reason free and open source software implementations such as FFmpeg and FAAC may be distributed in source form only, in order to avoid patent infringement. (See below under Products that support AAC, Software.) But the xSplit program had to cancel the AAC for free members because they have to pay royalties per person. Is this true (that you have to pay per each person that uses aac)? If you do have to pay, which company do you pay to and how does one apply?

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  • Free Mobile passe largement la barre des 3 millions d'abonnés et en vise trois fois plus sur le moyen terme

    Free lance le forfait mobile à 2€ Et l'illimité à 3Go par mois pour moins de 20€ Mise à jour du 10/01/12 « L'oligopole des opérateurs s'est entendu avec l'Etat sur le forfait RSA. En interne, nous on l'appelle le forfait arnaque-raquette ». Le ton est donné, Xavier Niel, PDG fondateur de Free, n'est pas là pour être diplomate. « Plus vous êtes pauvres, et plus on vous en met dans la tête ! ». Pour bien montrer que ces opérateurs « nous prennent pour des pigeons » et pour se placer en chevalier blanc sur un marché biaisé, Free prend l'exemple symbolique de ces abonnements « sociaux » au prix de 10€ par mois.

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  • Are there any free embedable REPL interpreters for a website? [migrated]

    - by Google
    I am hoping to find a free REPL interpreter for a number of languages. I am starting a new web page to help developers learn with a number of tutorials. I want to embed an interpreter on the site so users can have a tutorial and interpreter side by side. I really like free interpreters such as the one at repl.it. However, I am unable to find a free/embeddable interpreter. I was hoping to avoid making one myself-- they seem tricky to make safely! Thanks!

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  • How to build a private Wi-Fi Network server with VMware?

    - by Maarten Schermer
    For a school project, we have to build a Private network with VMware vSphere , which we can connect to with a Username and Password. On the network we want to create a folder for each Useraccount. Also we must have add a few groups (Admin,Customer,Manager). We must be able to connect to our network via the school Wi-Fi. We want to build a safe and secure network, with an easy way to access the network. Do you have any tips on how to approach this?

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  • How do I bridge a connection from Wi-Fi to TAP on Mac OS X? (for the emulator QEMU)

    - by penx
    I'm trying to setup a bridge between my Wi-Fi connection and an emulator (QEMU). I need a virtual machine to be on the same LAN as the host, with its own IP address. QEMU requires using a TAP (virtual network device) so I have installed tuntaposx, have it running, and can open up QEMU using a TAP: qemu-system-arm -kernel zImage.integrator -initrd arm_root.img -m 256 -net nic -net tap,ifname=tap1 -nographic -append "console=ttyAMA0" I have a script that configures the bridge once QEMU has opened up the TAP interface: sysctl -w net.link.ether.inet.proxyall=1 sysctl -w net.inet.ip.forwarding=1 sysctl -w net.inet.ip.fw.enable=1 ifconfig bridge0 create ifconfig bridge0 addm en1 ifconfig tap1 0.0.0.0 up ifconfig bridge0 addm tap1 ifconfig bridge0 up If I manually set an IP on the VM, I can ping from the VM to the host, but not from the host to the VM. Also, I can't access the rest of the network from the VM - including not being able to set an IP over DHCP. Any ideas?

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  • How can I make my laptop use a phone as a wi-fi adapter?

    - by Dennis Williamson
    I have an HP Pavilion dv6400 series laptop with an NVidea chipset which is the subject of a class action lawsuit. The symptom (PDF) that I'm experiencing is that the system fails to recognize that there is a wireless adapter installed. It doesn't appear in Device Manager. In this question, I ask if there's a way to work around this problem. As an alternative, I'd like to see if there's a way to use an old Windows Mobile 5 phone that does not have cellular service (voice or data) as a Wi-Fi adapter. How can this be accomplished? The system is running Windows Vista Home and has the latest BIOS (F42) and Windows updates and drivers.

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  • Is it better to use a crowded 2.4GHz Wi-Fi channel 1, 6, 11 or "unused" 3, 4, 8, or 9?

    - by Luke
    I understand that 2.4GHz Wi-Fi channels overlap, and that the only non-overlapping channels in the US are 1, 6, and 11. Generally, my signal strength on channels 1, 6, and 11 are much stronger than my neighbors' on the same channel. However, several of the channels may have 4 or 5 others on that same channel. In this scenario, is it better to use 3, 4, 8, or 9? Or is it better to use the crowded channels 1, 6, and 11? As a secondary question, does it even matter that my signal strength is much higher than theirs? Related: Why use wifi channels other than 1, 6 or 11?

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  • How can I offer 2.4 ghz wi-fi in a building with strong interference?

    - by user49995
    I have a few access points on one floor of a high-rise building. They support both 2.4 ghz and 5ghz. When I used 2.4 ghz, the channel management features did not seem to work and we experienced frequent problems. When I switched to 5 ghz the problems went away. However, the 5ghz standard is much less accepted. And when clients come in, they want 2.4ghz. What can I do? How can I offer 2.4ghz wi-fi in an area with a lot of interference?

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  • HP Presario CQ 61-322ER (VV884EA) Wi-Fi hang up! [closed]

    - by qgrabber
    Possible Duplicate: HP Presario CQ 61-322ER (VV884EA) Wi-Fi hang up! I have my new laptop and don't have Windows XP drivers for it. I found that it contains the Broadcom BCM4310 chip, but when I install any Broadcom driver my laptop hangup on installing bcm5*.sys driver. Only power-off button make any effect. After reboot the device list (Device Manager) contains Broadcom WLAN adapter, but it is marked as disabled, for some hardware errror! Also if I disable device before, and install driver - then - all is OK! But when I try to enable it, Windows hang up anyway (no speaker beep, no mouse input, no keyboard input - nothing) What is the solution?

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  • Grub2 : Windows 7 can't boot installing with Ubuntu 10.04 on different hard drive

    - by dellphi
    I use a dual boot with two hard disks and two OS is Ubuntu 10.04 and Windows 7. Windows 7 installed on the first disk, first partition. Grub is installed on a second hard disk MBR, and Ubuntu installed on an extended partition on a second hard drive. When I select Windows 7 on the Grub menu, the HDD lamp lights up briefly and then black screen on the monitor, with the status of the keyboard is still functioning. Until now (with the default boot from first HDD), I have to press F12 to get into the Grub to run Linux on a second HDD. ================ fdisk -l ================================ dellph1@dellph1-desktop:~$ fdisk -l omitting empty partition (5) Disk /dev/sda: 1000.2 GB, 1000204886016 bytes 255 heads, 63 sectors/track, 121601 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00087dec Device Boot Start End Blocks Id System /dev/sda1 * 1 23104 185582848+ 7 HPFS/NTFS /dev/sda2 23105 121601 791177122 5 Extended /dev/sda5 36107 74408 307660783+ 7 HPFS/NTFS /dev/sda6 74409 100081 206218341 7 HPFS/NTFS /dev/sda7 100082 121601 172859368+ 7 HPFS/NTFS Disk /dev/sdb: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x6d43dfb2 Device Boot Start End Blocks Id System /dev/sdb1 1 10030 80560066 5 Extended /dev/sdb5 * 1 5560 44657601 83 Linux /dev/sdb6 5560 9387 30736384 83 Linux /dev/sdb7 9387 10030 5164032 82 Linux swap / Solaris dellph1@dellph1-desktop:~$ ================= grub.cfg ================== # DO NOT EDIT THIS FILE # It is automatically generated by /usr/sbin/grub-mkconfig using templates from /etc/grub.d and settings from /etc/default/grub # BEGIN /etc/grub.d/00_header if [ -s $prefix/grubenv ]; then load_env fi set default="0" if [ ${prev_saved_entry} ]; then set saved_entry=${prev_saved_entry} save_env saved_entry set prev_saved_entry= save_env prev_saved_entry set boot_once=true fi function savedefault { if [ -z ${boot_once} ]; then saved_entry=${chosen} save_env saved_entry fi } function recordfail { set recordfail=1 if [ -n ${have_grubenv} ]; then if [ -z ${boot_once} ]; then save_env recordfail; fi; fi } insmod ext2 set root='(hd1,5)' search --no-floppy --fs-uuid --set 2f014a3a-35f3-4d05-87aa-34ca677160b7 if loadfont /usr/share/grub/unicode.pf2 ; then set gfxmode=1024x768 insmod gfxterm insmod vbe if terminal_output gfxterm ; then true ; else # For backward compatibility with versions of terminal.mod that don't # understand terminal_output terminal gfxterm fi fi insmod ext2 set root='(hd1,5)' search --no-floppy --fs-uuid --set 2f014a3a-35f3-4d05-87aa-34ca677160b7 set locale_dir=($root)/boot/grub/locale set lang=en insmod gettext if [ ${recordfail} = 1 ]; then set timeout=-1 else set timeout=5 fi END /etc/grub.d/00_header BEGIN /etc/grub.d/05_debian_theme insmod ext2 set root='(hd1,5)' search --no-floppy --fs-uuid --set 2f014a3a-35f3-4d05-87aa-34ca677160b7 insmod jpeg if background_image /usr/share/backgrounds/CurlsbyCandy.jpg ; then set color_normal=white/black set color_highlight=black/light-gray else set menu_color_normal=white/black set menu_color_highlight=black/light-gray fi END /etc/grub.d/05_debian_theme BEGIN /etc/grub.d/10_linux menuentry 'Ubuntu, with Linux 2.6.32-24-generic' --class ubuntu --class gnu-linux --class gnu --class os { recordfail insmod ext2 set root='(hd1,5)' search --no-floppy --fs-uuid --set 2f014a3a-35f3-4d05-87aa-34ca677160b7 linux /boot/vmlinuz-2.6.32-24-generic root=UUID=2f014a3a-35f3-4d05-87aa-34ca677160b7 ro splash vga=795 quiet splash nomodeset video=uvesafb:mode_option=1280x1024-24,mtrr=3,scroll=ywrap initrd /boot/initrd.img-2.6.32-24-generic } menuentry 'Ubuntu, with Linux 2.6.32-24-generic (recovery mode)' --class ubuntu --class gnu-linux --class gnu --class os { recordfail insmod ext2 set root='(hd1,5)' search --no-floppy --fs-uuid --set 2f014a3a-35f3-4d05-87aa-34ca677160b7 echo 'Loading Linux 2.6.32-24-generic ...' linux /boot/vmlinuz-2.6.32-24-generic root=UUID=2f014a3a-35f3-4d05-87aa-34ca677160b7 ro single splash vga=795 echo 'Loading initial ramdisk ...' initrd /boot/initrd.img-2.6.32-24-generic } END /etc/grub.d/10_linux BEGIN /etc/grub.d/30_os-prober menuentry "Windows 7 (loader) (on /dev/sda1)" { insmod ntfs set root='(hd0,1)' search --no-floppy --fs-uuid --set 5cac2139ac210f58 chainloader +1 } END /etc/grub.d/30_os-prober BEGIN /etc/grub.d/40_multisystem Ajout de MultiSystem MULTISYSTEM MENU menuentry "PLoP Boot Manager" { linux16 /boot/plpbt } menuentry "Smart Boot Manager" { search --set -f /boot/sbootmgr.dsk linux16 /boot/memdisk initrd16 /boot/sbootmgr.dsk } FIN MULTISYSTEM MENU END /etc/grub.d/40_multisystem ================================================ I want to keep the Grub on the second HDD. I have been using the Startup Manager, Boot Manager and Grub Customizer, and this problem still unsolved. The easiest thing that I can possibly do is to install Grub on first HDD, but I was curious and maybe someone can help.

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  • Shift-reduce: when to stop reducing?

    - by Joey Adams
    I'm trying to learn about shift-reduce parsing. Suppose we have the following grammar, using recursive rules that enforce order of operations, inspired by the ANSI C Yacc grammar: S: A; P : NUMBER | '(' S ')' ; M : P | M '*' P | M '/' P ; A : M | A '+' M | A '-' M ; And we want to parse 1+2 using shift-reduce parsing. First, the 1 is shifted as a NUMBER. My question is, is it then reduced to P, then M, then A, then finally S? How does it know where to stop? Suppose it does reduce all the way to S, then shifts '+'. We'd now have a stack containing: S '+' If we shift '2', the reductions might be: S '+' NUMBER S '+' P S '+' M S '+' A S '+' S Now, on either side of the last line, S could be P, M, A, or NUMBER, and it would still be valid in the sense that any combination would be a correct representation of the text. How does the parser "know" to make it A '+' M So that it can reduce the whole expression to A, then S? In other words, how does it know to stop reducing before shifting the next token? Is this a key difficulty in LR parser generation?

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  • ANTLRv3 How to set a custom Lexer and Parser Class names

    - by Filip
    Is there a way to specify custom class name (meaning independent of the grammar name) for the ANTLRv3 generated Parser and Lexer classes? So in the case of grammar MDD; //other stufff Automtaically it would crate MDDParser and MDDLexer, but i would like to have them for example as MDDBaseParser and MDDLexer.

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  • Best Open Source Project Hosting Site

    - by Cristian
    I want to start an open source project, but the rise in hosting sites leaves me a little paralyzed with choice. I know a little about several: I never really liked SourceForge's UI but it still feels like the site I think of when I think "open source project hosting". Google Code Project Hosting looks clean and useful but doesn't seem as feature complete as SourceForge. I've heard good things about Launchpad but don't know much about it nor do I know Bazaar (though I'd be interested in learning it). I know almost nothing about GitHub and, like Bazaar, I don't know Git. Does anyone have any experience with these sites or some other cool code host? Any recommendations? Recommended Sites: BitBucket Codeplex Assembla DevjaVu Savannah

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  • Commercial use of open source software

    - by gsvirdi
    I'd been trying to read about Open Source Software (OSS) but I'm still not clear about the scope of commercial use of OSS. Where can I find more details about the commercial use of OSS? Only then can I concentrate on "What License should I use for a new OSS project in .NET" for FOSS.

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  • Understanding CLR 2.0 Memory Model

    - by Eloff
    Joe Duffy, gives 6 rules that describe the CLR 2.0+ memory model (it's actual implementation, not any ECMA standard) I'm writing down my attempt at figuring this out, mostly as a way of rubber ducking, but if I make a mistake in my logic, at least someone here will be able to catch it before it causes me grief. Rule 1: Data dependence among loads and stores is never violated. Rule 2: All stores have release semantics, i.e. no load or store may move after one. Rule 3: All volatile loads are acquire, i.e. no load or store may move before one. Rule 4: No loads and stores may ever cross a full-barrier (e.g. Thread.MemoryBarrier, lock acquire, Interlocked.Exchange, Interlocked.CompareExchange, etc.). Rule 5: Loads and stores to the heap may never be introduced. Rule 6: Loads and stores may only be deleted when coalescing adjacent loads and stores from/to the same location. I'm attempting to understand these rules. x = y y = 0 // Cannot move before the previous line according to Rule 1. x = y z = 0 // equates to this sequence of loads and stores before possible re-ordering load y store x load 0 store z Looking at this, it appears that the load 0 can be moved up to before load y, but the stores may not be re-ordered at all. Therefore, if a thread sees z == 0, then it also will see x == y. If y was volatile, then load 0 could not move before load y, otherwise it may. Volatile stores don't seem to have any special properties, no stores can be re-ordered with respect to each other (which is a very strong guarantee!) Full barriers are like a line in the sand which loads and stores can not be moved over. No idea what rule 5 means. I guess rule 6 means if you do: x = y x = z Then it is possible for the CLR to delete both the load to y and the first store to x. x = y z = y // equates to this sequence of loads and stores before possible re-ordering load y store x load y store z // could be re-ordered like this load y load y store x store z // rule 6 applied means this is possible? load y store x // but don't pop y from stack (or first duplicate item on top of stack) store z What if y was volatile? I don't see anything in the rules that prohibits the above optimization from being carried out. This does not violate double-checked locking, because the lock() between the two identical conditions prevents the loads from being moved into adjacent positions, and according to rule 6, that's the only time they can be eliminated. So I think I understand all but rule 5, here. Anyone want to enlighten me (or correct me or add something to any of the above?)

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  • Why does ANTLR not parse the entire input?

    - by Martin Wiboe
    Hello, I am quite new to ANTLR, so this is likely a simple question. I have defined a simple grammar which is supposed to include arithmetic expressions with numbers and identifiers (strings that start with a letter and continue with one or more letters or numbers.) The grammar looks as follows: grammar while; @lexer::header { package ConFreeG; } @header { package ConFreeG; import ConFreeG.IR.*; } @parser::members { } arith: term | '(' arith ( '-' | '+' | '*' ) arith ')' ; term returns [AExpr a]: NUM { int n = Integer.parseInt($NUM.text); a = new Num(n); } | IDENT { a = new Var($IDENT.text); } ; fragment LOWER : ('a'..'z'); fragment UPPER : ('A'..'Z'); fragment NONNULL : ('1'..'9'); fragment NUMBER : ('0' | NONNULL); IDENT : ( LOWER | UPPER ) ( LOWER | UPPER | NUMBER )*; NUM : '0' | NONNULL NUMBER*; fragment NEWLINE:'\r'? '\n'; WHITESPACE : ( ' ' | '\t' | NEWLINE )+ { $channel=HIDDEN; }; I am using ANTLR v3 with the ANTLR IDE Eclipse plugin. When I parse the expression (8 + a45) using the interpreter, only part of the parse tree is generated: http://imgur.com/iBaEC.png Why does the second term (a45) not get parsed? The same happens if both terms are numbers. Thank you, Martin Wiboe

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  • What does the q in a q-grammar stand for?

    - by Aru
    So I've been reading sites and the classic books on compilers, reading about s-grammar and q-grammars I wondered what the s and q stand for, I think the s stands for simple grammar. While the q...well, I have no idea. What does the q in a q-grammar stand for?

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  • Wikipedia article's

    - by Algorist
    Hi, I am doing a project, for which I need to know all the wikipedia article names(I don't need the content). Is there a place where I can download this data. Thank you Bala

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  • Java: how to tell if a line in a text file was supposed to be blank?

    - by defn
    I'm working on a project in which I have to read in a Grammar file (breaking it up into my data structure), with the goal of being able to generate a random "DearJohnLetter". My problem is that when reading in the .txt file, I don't know how find out whether the file was supposed to be a completely blank line or not, which is detrimental to the program. Here is an example of part of the file, How do i tell if the next line was supposed to be a blank line? (btw I'm just using a buffered reader) Thanks! <start> I have to break up with you because <reason> . But let's still <disclaimer> . <reason> <dubious-excuse> <dubious-excuse> , and also because <reason> <dubious-excuse> my <person> doesn't like you I'm in love with <another> I haven't told you this before but <harsh> I didn't have the heart to tell you this when we were going out, but <harsh> you never <romantic-with-me> with me any more you don't <romantic> any more my <someone> said you were bad news

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  • Recognizing terminals in a CFG production previously not defined as tokens.

    - by kmels
    I'm making a generator of LL(1) parsers, my input is a CoCo/R language specification. I've already got a Scanner generator for that input. Suppose I've got the following specification: COMPILER 1. CHARACTERS digit="0123456789". TOKENS number = digit{digit}. decnumber = digit{digit}"."digit{digit}. PRODUCTIONS Expression = Term{"+"Term|"-"Term}. Term = Factor{"*"Factor|"/"Factor}. Factor = ["-"](Number|"("Expression")"). Number = (number|decnumber). END 1. So, if the parser generated by this grammar receives a word "1+1", it'd be accepted i.e. a parse tree would be found. My question is, the character "+" was never defined in a token, but it appears in the non-terminal "Expression". How should my generated Scanner recognize it? It would not recognize it as a token. Is this a valid input then? Should I add this terminal in TOKENS and then consider an error routine for a Scanner for it to skip it? How does usual language specifications handle this?

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  • Theory of computation - Using the pumping lemma for CFLs

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

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  • Freeing memory twice

    - by benjamin button
    Hi, AFAIK, freeing a NULL pointer will result in nothing. I mean nothing is being done by the compiler/no functionality is performed. Still, I do see some statements where people say that one of the scenarios where memory corruption can occur is "freeing memory twice"? Is this still true?

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