I mistakenly forgot to include the name of a function when writing a statement. Only in debugging did I find this out as everything compiled fine. I am surprised that gcc did not produce a warning for a C program like this.  int x =0, a=1, b=2, c=3, d=4, e=5, f=6, g=7;      printf("\n%d %d %d %d %d %d %d %d\n",x,a,b,c,d,e,f,g);  x = (a,g=b+c,d,e,f);      printf("\n%d %d %d %d %d %d %d %d\n",x,a,b,c,d,e,f,g);  x = a,g=b+c,d,e,f;      printf("\n%d %d %d %d %d %d %d %d\n",x,a,b,c,d,e,f,g);   The code should have read functionName(a,g=b+c,d,e,f) but was missing the function name.  The syntax  x = (a,g=b+c,d,e,f) results in x = f The syntax  x = a,g=b+c,d,e,f   results in x = a  In both cases g was properly calculated as b+c.  Wouldn't a warning of the first version x = (...) been in order? What valid stand alone expression would appear with (...) in this way. The byproduct is the inclusion or exclusion of the parens also results in two different values for x. Setting x=f due to the parens looks like a fall out from the way the machine code was generated as opposed to being syntactically driven.   Seems that a warning should have been flagged that perhaps this grouped expression was really the parameters of a missing function call. Better to catch a warning like this than having to discover the issue during debugging.
What does the , operator do in C?
My chooser (file-picker dialog) goes under the dock so I can't see buttons. What can I do? jQuery select and wrap textNode how to remove the double quotes in a csv I have a CSV file like  a.csv "1,2,3,4,9" "1,2,3,6,24" "1,2,6,8,28" "1,2,4,6,30"   I want something like   b.csv 1,2,3,4,9 1,2,3,6,24 1,2,6,8,28 1,2,4,6,30   I tried awk '{split($0,a,"\""); But did not help.Any help is appreciated. Why does the SARS-CoV2 virus not remain infectious forever? Or does it? Given that the majority of biologists do not currently consider viruses to be alive, a virus can never die.  It can, however, get destroyed by long exposures to soapy water, alcohol, and apparently certain frequencies of UV light.  Why is an individual SARS-CoV2 virion not infectious forever?  Or is it? Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$?  Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing. When saving a .qgs file in QGIS, another appears. Is this a backup? When saving a project file (.qgs), which when executed is automatically loaded with qgis.bat, there's another .qgs file which appears in the same directory of the original project file, although this one is not loaded with the qgis.bat file:    This is just out of curiosity but is the latter designed to be a backup of the original .qgs file? How can a function with a hole (removable discontinuity) equal a function with no hole? I've done some research, and I'm hoping someone can check me. My question was this:  Assume I have the function $f(x) = \frac{(x-3)(x+2)}{(x-3)}$, so it has removable discontinuity at $x = 3$. We remove that discontinuity with algebra: $f(x) = \frac{(x-3)(x+2)}{(x-3)} = (x+2)$. BUT, the graph of the first function has a hole at $x = 3$, and the graph of the second function is continuous everywhere. How can they be "equal" if one has a hole and the other does not?  I think that this is the answer:   Because the original function is undefined at the point $x = 3$, we have to restrict the domain to $\mathbb{R} \setminus 3$. And when we manipulate that function with algebra, the final result, $f(x) = (x + 2)$ is still using this restricted domain. So even though the function $f(x) = (x+2)$ would not have a hole if the domain were all of $\mathbb{R}$, we are sort of "imposing" a hole at $x = 3$ by continuing to throw that point out of the domain.   And then just to close the loop: Removing the removable discontinuity is useful because it allows us to "pretend" that we're working with a function that is everywhere continuous, which helps us easily find the limit. But the reality is that the function $f(x) = (x +2)$ is actually NOT continuous everywhere when we restrict the domain by throwing out the point 3. Or am I now taking things too far?   Thanks in advance!  EDIT: For anyone coming across this in the future, in addition to the excellent answers below, I also found this other question about the  helpful. I'm porting a module that currently works with Blocks/BlockForms to Paragraphs/NodeForms. I'm inserting an embedded form inside a Paragraph inside a Node Edit page using a hook_field_widget_entity_reference_paragraphs_form_alter. The general structure where the form gets inserted is: $form['field']['widget'][0]['settings'][form here]  After this the form displays as expected on the node edit page. The inserted form goes through a number of #process steps. The element in question is a select form that when the button underneath it is pressed, is supposed to retrieve some data via a custom submit handler and place it on the page with ajax:  The array for this element looks like this:  You can see the custom submit handler that is supposed to fire when 'Select Segment Set' is pressed After debugging, I've found that the entire NodeForm appears to be submitted to the backend, and the only handler that is found is from some other field that uses ajax. In short 'updateSegmentSet' is never discovered as a submit handler. When I debug the original module based on block/blockforms, I do see the correct submit handler being discovered. Now in the latter case, of course a blockForm is submitted to the back end with the correct plugin. Given that this is basically a form within a form within a form, is there something special I have to do to get the custom submit handler to run? I've read a few things online e.g. the 'settings' key and '#tree' = true. EDIT The Ajax property looks like this:       '#ajax' =&gt; [         'callback' =&gt; [$this, 'updateSegmentSetAjax'],         'wrapper' =&gt; $wrapper_id,       ],    public function updateSegmentSetAjax(array &amp;$form, FormStateInterface $form_state) {     $button = $form_state-&gt;getTriggeringElement();     // Go one level up in the form, to the widgets container.     return NestedArray::getValue($form, array_slice($button['#array_parents'], 0, -2));   }  As best I can tell, this is never called either. Calculate the limit including binomial coefficients Quantum Entanglement - What's the big deal? Bearing in mind I am a layman - with no background in physics - please could someone explain what the "big deal" is with quantum entanglement?  I used to think I understood it - that 2 particles, say a light-year apart spatially, could affect each other physically, instantly.  Here I would understand the "big deal".  On further reading I've come to understand (maybe incorrectly) that the spatially separated particles may not affect each other, but in knowing one's properties you can infer the other's.  If that it the case, I don't see what the big deal is... 2 things have some properties set in correlation to each other at the point of entanglement, they are separated, measured, and found to have these properties...?  What am I missing? Is it that the particles properties are in an "un-set" state, and only when measured do they get set?  (i.e. the wave-function collapses).  If this is true - why do we think this instead of the more intuitive thought that the properties were set at an earlier time? How can I add vertices to intersection of two edges? I need to put a point at the intersection of two edges. Below is a picture with red circles around where I need points to be placed. I need to find limit points of the set $\{  \frac{1}{n} +\frac{1}{m} \mid n, m = 1,2,3,\dots \}$.  My try :   If both $m$ and $n$ tend to very large values say $\infty$ then the value of $\{ \frac{1}{n} +\frac{1}{m} \}$ tends to $0$, and if only one of $m$ or $n$ tends to very large values, then the set $\{ \frac{1}{n} \mid n=1,2,3,\dots\} $ acts as limit points.   So is it true that a set of limit points is $\{0\}$ $\cup$  $\{\frac{1}{k} \mid k=1,2,3,\dots \}$?  How should I write this proof rigorously? Geocoding USA addresses that cannot be sent over internet? For a one-time project, I need to geocode a few thousand addresses.  In the past I've used various online resources for this kind of thing (e.g., Google Maps API), but the addresses I'm working with have to be kept confidential - which means no sending it over the Internet, unless there's some iron-clad guarantee of privacy.  What other options do I have? Which of these are acceptable? Is one preferable over the other?   "Chemically-deposited tourmaline is never periwinkle." "Chemically deposited tourmaline is never periwinkle."   Also, is the title to the this question asking what I'm trying to ask?  (I realize that there are other questions about hyphenation in similar contexts; I'm asking specifically about -ly adverbs.) What is the best way to clean the sensor on a interchangeable lens camera (mirrorless or digital SLR)? If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If  $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $. Progress We have   $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$. Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$. How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $. How do I register a custom URL protocol in Windows? How do I register a custom protocol with Windows so that when clicking a link in an email or on a web page my application is opened and the parameters from the URL are passed to it? Additional needed documents for UK trip if someone will shoulder all expenses Someone (a relative) from the UK (already a naturalized British Citizen) will be financing my trip, as well as taking care of my accommodation. What are the documents I need to submit aside from bank statements? Does the sponsor need to fill-out the Sponsorship Undertaking Form? ()  something to consider from my question: -my relative is a naturalized citizen in the UK. Is proof of naturalization needed? How to upgrade from 18.10 to 19.10, using the command line? I'm trying to upgrade from 18.10 to 19.10, using sudo do-release-upgrade however I only get an error: An upgrade from 'cosmic' to 'eoan' is not supported with this tool. I performed the steps to update my sources.list to point to old-release server, as suggested here:  This did not help. I get the same error. Can negative levels get rid of racial HD? What happens when one is affected by something such as energy drain, and permanently loses levels when one only has racial HD? Does the victim of the spell die? Or do they permanently lose racial HD?