Can Somebody travel with their gun internationally?
Carrying a handgun in other countries with a US concealed carry permit
Is there a way to configure your Stack Overflow account so that you receive notification emails when posts are made to specific tags?    Or do you simply have to log on to see new posts? Nth max salary in Oracle In chapter 2, the section on bitwise operators (section 2.9), I'm having trouble understanding how one of the sample methods works.  Here's the method provided:  unsigned int getbits(unsigned int x, int p, int n) {     return (x &gt;&gt; (p + 1 - n)) &amp; ~(~0 &lt;&lt; n); }   The idea is that, for the given number x, it will return the n bits starting at position p, counting from the right (with the farthest right bit being position 0). Given the following main() method:  int main(void) {     int x = 0xF994, p = 4, n = 3;     int z = getbits(x, p, n);     printf("getbits(%u (%x), %d, %d) = %u (%X)\n", x, x, p, n, z, z);      return 0; }   The output is:     getbits(63892 (f994), 4, 3) = 5 (5)   I get portions of this, but am having trouble with the "big picture," mostly because of the bits (no pun intended) that I don't understand.  The part I'm specifically having issues with is the complements piece: ~(~0 &lt;&lt; n). I think I get the first part, dealing with x; it's this part (and then the mask) that I'm struggling with -- and how it all comes together to actually retrieve those bits. (Which I've verified it is doing, both with code and checking my results using calc.exe -- thank God it has a binary view!)  Any help? Highest power of a prime $p$ dividing $N!$ Is my MAC address public when browsing the Internet? How do I forward/NAT all traffic to one interface/IP to a remote IP? I have one "server A" that has multiple IPs attached to it, like so:  eth0:0 1.1.1.1 eth0:1 1.1.1.2 eth0:2 1.1.1.3   I have another "server B" that also has multiple IPs attached to it, like so:  eth0:0 2.2.2.1 eth0:1 2.2.2.2 eth0:2 2.2.2.3   Now, I want to setup iptables on "Server A" to forward/NAT all incomming traffic on "eth0:2" to IP 2.2.2.3 on "Server B".  I have verified that "Server A" is able to "talk" to "Server B" on IP 2.2.2.3. Ping and telnet to open ports works just fine and I have the forward-flag turned on (net.ipv4.ip_forward=1)  I have tried multiple different ways, DNAT, SNAT, MASQUERADE etc, but I cannot get anything to work.  This line works fine if I try to forward traffic between IPs on the SAME server:  iptables -t nat -A PREROUTING -d 1.1.1.3 -j DNAT --to-destination 1.1.1.2   But if I switch out the "1.1.1.2" for "2.2.2.3", it does not work.  I assume that I need a second iptable rule to solve it. I have tried with the following POSTROUTING rules without any luck (not at the same time):  iptables -t nat -A POSTROUTING -d 2.2.2.3 -j MASQUERADE iptables -t nat -A POSTROUTING -d 2.2.2.3 -j SNAT --to 1.1.1.3 iptables -t nat -A POSTROUTING -j MASQUERADE   What am I missing?  EDIT 1:  I finally got it to work by using the following:  net.bridge.bridge-nf-call-iptables=0  iptables -t nat -A PREROUTING -d 1.1.1.3 -j DNAT --to-destination 2.2.2.3 iptables -t nat -A POSTROUTING -d 2.2.2.3 -j SNAT --to 1.1.1.3   However, now another problem appeared. All logs etc on server 2.2.2.3 shows that ALL traffic now comes from 1.1.1.3, like apache logs, mail logs etc. I assume this is the nature of NAT.   However, when I do standard port forwarding on my home-router to my laptop that is running apache, I see the original "requester IP" in the logs. So, how does the router do this? And how can I do the same on my server setup?  Bottom line, I want to forward all traffic from Server A (1.1.1.3) to Server B (2.2.2.3), BUT I also want to be able to see where the traffic came from on Server B (2.2.2.3), i.e the apache logs should show the original IP of the requester.  I assume I should use some other way than NAT to make this happen, and it should be possible, as even my simple home-router is able to do this.  One extra thing, the IPs attached to Server A and Server B is LOCKED to each respective server. Thus, Server A is NOT able to send out traffic FROM IP 2.2.2.3. It is locked by my provider in the router. $a\mid b,\ c\mid d\,\Rightarrow\ ac\mid bd $ $\ \, \bf\small [Divisibility\ Product\ Rule]$ I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$. There have been  for a while, but this is ridiculous. It reminds me of , but this time it's network-wide.   The banner must be scrolled past (or dismissed) on every page unless both first-party JavaScript and first-party cookies are enabled. If first-party JavaScript is enabled, and the banner is visible, the rest of the page jumps up and down every so often.  Under certain circumstances (not quite sure what they are, yet – a direct link to an answer, maybe? I'm seeing it ) the entire page jumps up and down a line every few seconds even when the banner is off-screen.   The cookie banner keeps coming back unless both third-party JavaScript  are enabled (and  is solved). The cookie box occludes the “Learn more about Teams” link in the banner (with default Firefox window sizes) and the Job / Teams link in the sidebar.  As it stands,  – a site for hobbyists and researchers, whose audience is not very likely to purchase Teams – is much, much harder to use, for no benefit that I can see. Can you roll back this change, please? Logistic regression with binary dependent and independent variables How would I have a  action that may have some effects on the current page but would also change the URL in the browser so if the user hits reload or bookmark, then the new URL is used?    It would also be nice if the back button would reload the original URL.  I am trying to record JavaScript state in the URL. What if a light clock travels perpendicular to mirrors that make up the clock? Locking an iPhone if button is broken? My friend's iPhone's sleep/wake button no longer works. Is there any way to turn off the screen besides setting up auto-lock and waiting a few minutes? I've always been taught to put myself last when referring to myself in the same sentence as others but the usage of "me and..." seems to be everywhere these days. The misuse of the word "me" instead of "I" aside, is there some new rule I haven't heard of? Shouldn't we put ourselves last regardless of the "me"/"I" usage?  Examples of "correct" usage:      My friends and I went for some ice   cream. Did you see my friends and me   at the ice cream stand?   Examples of "incorrect" usage:     Me and my friends went for some ice   cream. Did you see me and my friends   at the ice cream stand?   Note: I was also taught that the only person who could put themselves first was the queen. I just fit a model in lme4, and I'm wondering what the heck I fit...  I have individuals id, and each is measured pass/fail on items that can be described using two factors, f1 and f2. My theory says that id and f2 can interact. So I want to compare these models:  # f2 doesn't matter m1 &lt;- lmer(pass ~ f1 + (1|id), family="binomial")  # f2 is a fixed effect, doesn't interact with individuals m2 &lt;- lmer(pass ~ f1 + f2 + (1|id), family="binomial")  # f2 is a nested random effect, interacting with individuals m3 &lt;- lmer(pass ~ f1 + (1|f2/id), family="binomial")   First, am I right that there's a random effect of each level of f2, and each individual's level of f2 is partially pooled, depending on the number of observations for that individual?  Second, does it make sense to fit f2 as a fixed effect with random individuals nested within each level of f2? How does one write that?  Third, what's this, with f2 and id reversed in the formula? I typed it by accident, but now I want to understand it.  m3weird &lt;- lmer(pass ~ f1 + (1|id/f2), family="binomial") Two definitions of $\limsup$ Group of $r$ people at least three people have the same birthday? What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday?   $\displaystyle 1- \frac{365\cdot364 \cdots(365-r+1)}{365^r}$   $\displaystyle \frac{365\cdot364         \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots         (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle 1- \frac{365\cdot364             \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle\frac{365\cdot364 \cdots(365-r+1)}{365^r}                 $     My attempt :  (May be typo in option $(3)$ of question !)  $$P(\text{at least 3 persons have same birthday})$$  $$= 1 - \{P\text{(no one has same birthday) + P(any 2 have same birthday)\}}$$  So, option $(3)$ is true.     Can you explain it, please?   It  before,  but I'm not satisfied by explanation. error in count script for number of occurences I just signed up, a day or two ago, and I've received a notification that there's something in my inbox.  Unfortunately, my email client isn't able to show me a link to my Inbox, or the message, because of the format of the email.  I normally use the email address that I gave when I signed up, just with MailWasher, which is intended to safely preview emails, as plain text, to decide whether or not to open them in a normal HTML email client.  I would be nice if that email contained a link to my inbox, which could be shown in MailWasher.  In order to collect that email in a normal email client, I'd also have to collect or delete all the other emails sent to that address.  I'm not entirely clear what's been done in that email to stop MailWasher seeing the link.  I can see a link to leave feedback, and one to unsubscribe, but not one to check my inbox.  Sure, I can live without one, but those emails would be a lot more useful to me if I could just click a link to go to my Inbox. Drawing cross section in QGIS? Does anyone have a simple work-flow for drawing cross sections of earth features in QGIS?   My thing is that QGIS is being used to present the plan view, but we have a little interest in some elevations as well. All the data is one set and is simply an elevation value in the attribute table, I don't want to have to move all that to CAD.  One logical workaround is to present the z value as the y.  I am a dunce in paper space in CAD, and would be more comfortable with the QGIS Composer. A site URL changed and now its icon in SEDE is broken