GPS parser function doesn’t work with new receiver

I’m using an old device (2006 – extremely trimmed embedded linux) to get Gps latitude and longitude from an external receiver. Gps coordinates are displayed on device’s LCD display. It used to work fine until the original Gps receiver failed. They replaced it with a new receiver but now it doesn’t work anymore, all I see on LCD is: Latitude: – / Longitude: -.

I tried to see if the new Gps receiver is operational. And it seems the GPGGA packets are there. The old device expects GPGGA packets for parsing/processing.

char GPSLatBuf[12], GPSLonBuf[12];  int parseGPSData(char * gpsBuffer) {   char * p;   char gpsQuality = '0';    if((p = strstr(gpsBuffer, "$  GPGGA")) == NULL)      return -1;    memset(GPSLatBuf, 0, sizeof(GPSLatBuf));   memset(GPSLonBuf, 0, sizeof(GPSLonBuf));    p += 7; // UTC time   if(*p != ',')    {      p += 10; // Latitude      if(*p != ',')       {         memcpy(&GPSLatBuf[1], p, 9);         p += 10; // N/S Hemisphere         if(*p != ',')          {            GPSLatBuf[0] = (*p == 'N') ? '+' : '-';            p += 2; // Longitude            if(*p != ',')             {               memcpy(&GPSLonBuf[1], p, 10);               p += 11; // E/W Hemisphere               if(*p != ',')                {                  GPSLonBuf[0] = (*p == 'E') ? '+' : '-';                  p += 2; // GPS quality                  if(*p != ',')                     gpsQuality = *p;               }            }         }      }   }   if(gpsQuality == '0')      return -1;    return 0; } 

Expected to see non-void fields like before: +4916.4600 / -12311.1200 (just an example as I see nothing on display right now).

Why does this trick to derive the formula for $[A^n,B]$ in terms of repeated commutators work so well?

It is a known result that, given generically noncommuting operators $ A,B$ , we have $ $ A^n B=\sum_{k=0}^n \binom{n}{k} \operatorname{ad}^k(A)(B) A^{n-k}.\tag A $ $ This can be proved for example via induction with not too much work.

However, while trying to get a better understanding of this formula, I realised that there is a much easier way to derive it, at least on a formal, intuitive level.

The trick

Let $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ (standing for “shift” and “commute”, respectively) denote operators that act on expressions of the form $ A^k D^j A^\ell$ (denoting for simplicity $ D^j\equiv\operatorname{ad}^j(A)(B)$ ) as follows:

\begin{align} \hat{\mathcal S} (A^k D^j A^\ell) &= A^{k-1} D^j A^{\ell+1}, \ \hat{\mathcal C} (A^{k-1} D^{j+1} A^\ell) &= A^{k-1} D^j A^{\ell+1}. \end{align} In other words, $ \hat{\mathcal S}$ “moves” the central $ D$ block on the left, while $ \hat{\mathcal C}$ makes it “eat” the neighboring $ A$ factor.

It is not hard to see that $ \hat{\mathcal S}+\hat{\mathcal C}=\mathbb 1$ , which is but another way to state the identity $ $ A[A,B]=[A,B]A+[A,[A,B]].$ $ Moreover, crucially, $ \hat{\mathcal S}$ and $ \hat{\mathcal C}$ commute. Because of this, I can write

$ $ A^n B=(\hat{\mathcal S}+\hat{\mathcal C})^n (A^n B)=\sum_{k=0}^n\binom{n}{k} \hat{\mathcal S}^{n-k} \hat{\mathcal C}^{k}(A^n B),$ $ which immediately gives me (A) without any need for recursion or other tricks.

The question

Now, this is all fine and dandy, but it leaves me wondering as to why does this kind of thing work? It looks like I am somehow bypassing the nuisance of having to deal with non-commuting operations by switching to a space of “superoperators”, in which the same operation can be expressed in terms of commuting “superoperators”.

I am not even sure how one could go in formalising this “superoperators” $ \hat{\mathcal S},\hat{\mathcal C}$ , as they seem to be objects acting on “strings of operators” more than on the elements of the operator algebra themselves.

Is there a way to formalise this way of handling the expressions? Is this a well-known method in this context (I had never seen it but I am not well-versed in this kinds of manipulations)?

Unexpected document received html document SharePoint Work Bench

I have configured CMIS in SharePoint Few Sub Sites.

When I access particular Site CMIS Work Bench Throws an error

Unexpected document received: html document

I have another subsite(In same site Collection) wich is working perfectly with CMIS Work bench.

enter image description here

Thank you

Note: CMIS feature(Site Features) and Basic Authentication(Web Application) are Enabled.

Session d’osent work well in Drupal 8

I work on e-commerce web cite using drupal 8, and i use session To store items in basket. in local everythink is fin butin prod enivrement session dosent work

i used many implementation of session :

//1 $  tempstore = \Drupal::service('user.private_tempstore') //2 $  tempstore = \Drupal::service('user.private_tempstore') //3 use Symfony\Component\HttpFoundation\Session\Session; $  session = new Session(); //4 $  session_manager = \Drupal::service('session_manager'); 

but no solution seccees

Can A53 work with pl310?

I didn’t find any information that any A53 platform is working with pl310 cache contronller.

I’m not sure using cache controller like pl310 is still a typical way for A53 aarch64 platform.

If I have to use pl310 in my A53, how can I write driver for it under Linux? Lots of pl310 registers are “secure” write only.

Can a bard10/sorcerer1 take 6th level spells for magical secrets because of the way spell caster’s slots work? D&D5E [duplicate]

This question already has an answer here:

  • Can you use Magical Secrets to choose high level spells when multiclassing as a Bard? 2 answers

I have a lore bard who multi-classed into sorcerer after level 4 and went back to leveling as a bard. When he reaches 10th level bard (11th level spell caster) he will have 6th level slots. Can he take 6th level magical secrets because of this? The PHB states “Choose two Spells from any class, including this one. A spell you choose must be of a level you can cast, as shown on the Bard table, or a cantrip.” but the way multi-classing works means that he no longer uses the bard table and uses the spell caster multi-class table instead, right?

How to work around the IMPORTRANGE error: “Results too large”?

I’m trying to IMPORTRANGE from a range containing 240,000 cells (40 columns and 6,000 rows). The IMPORTRANGE function is erroring out with “Results too large”. I can’t seem to find documentation on the function’s limitations.

What are the limitations of IMPORTRANGE?

How can I get around this so I can import that data into my sheet?

Google Chrome: works. nslookup: work. Safari: does not work. ping: does not work

So I’m on a hotel wifi that has been working fine for several days.

Suddenly, I’m not able to use many programs on the internet.

Safari says “Can’t find server”.

Ping says “Can’t resolve: Unknown host”.

Meanwhile, I can browse just fine with Google Chrome and nslookup immediately returns correct results.

I’ve tried:

sudo dscacheutil -flushcache
sudo killall -HUP mDNSResponder

No change.

This has happened to me before, but usually the cycling the WiFi, or the commands above or a reboot would fix the problem…

Attempting to flash TWRP on Samsung Galaxy Tab S2 8.0 simply doesn’t work

I’m attempting to flash and boot into TWRP to install LineageOS, but try as I might the process described on the LOS website simply doesn’t work. My first step is to attempt to flash TWRP via Heimdall. I boot into Download Mode, and run the following command:

 heimdall flash --RECOVERY twrp-3.2.3-0-gts28vewifi.img --no-reboot 

It’s not really clear whether this works. Heimdall claims it was 100% successful, but a progress bar shows up with what looks like 1% remaining, that never completes. If I shut down here and attempt to reboot into recovery mode by holding Power + Volume Up + Home, the device just hangs at the splash screen that says Samsung Galaxy. I’ve tried literally dozens of different guides and nothing works.