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• What is wrong with this code for reading binary files? [on hold]

  - by qed

  What is wrong with this code for reading binary files? It compiles OK, but will not print out the file as planned, in fact, it prints nothing at all. #include <iostream> #include <fstream> int main(int argc, const char *argv[]) { if (argc < 2) { ::std::cerr << "usage: " << argv[0] << " <filename>\n"; return 1; } ::std::basic_ifstream<unsigned char> in(argv[1], ::std::ios::binary); unsigned char uc; while (in.get(uc)) { printf("%02X ", uc); } // TODO: error handling, in case the file could not be opened or read return 0; }

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• How to access stdClass variables stdClass Object([max(id)])=>64)

  - by Theopile

  I need the very last valid entry in a database table which would be the row with the greatest primary key. So using mysqli, my query is "SELECT MAX(id) FROM table LIMIT 1". This query returns the correct number(using print_r()) but I cannot figure out how to access it. Here is the main code. Note that the $this-link refers to class with a mysqli connection. $q="select max(id) from stones limit 1"; $qed=$this->link->query($q) or die(mysqli_error()); if($qed){ $row=$qed->fetch_object(); print_r($row); echo $lastid=$row;//here is the problem } The valid line print_r($row) echos out "stdClass Object ( [max(id)] = 68 )"

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• using an already proved lema/theorem/corollary in coq

  - by André Hincu

  I am trying to make a proof in Coq, and I would like to use a lemma already definded and proved by me. Is it possible for the following code? Lemma conj_comm: forall A B : Prop, A /\ B -> B /\ A. Proof. intros. destruct H. split. exact H0. exact H. Qed. Lemma not_conj_comm: forall A B : Prop, ~(A /\ B) -> ~(B /\ A). Proof. intros. intro. unfold not in H. apply H. use H0. In the above I want to use the fact that A /\B is the same as B /\ A in order to prove that ~(A /\ B) is the same as ~(B /\ A). Is it possible to use my proved lemma?

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• MatheMagics - Guess My Age Method 1

  - by PointsToShare

  © 2011 By: Dov Trietsch. All rights reserved MatheMagic – Guess My Age – Method 1 The Mathemagician stands on the stage and asks an adult to do the following: ·         Do the next few steps on your calculator, or the calculator in your phone, or even on a piece of paper. ·         Do it silently! Don’t tell me the results until I ask for them directly ·         Compute a single digit multiple of 9 – any one of 9, 18, 27, … all the way to 81, will do. ·         Now multiply your age by 10 ·         Subtract the 9 multiple from this number. ·         Tell me the result. Notice that I don’t know which multiple of 9 you subtracted from 10 times your age. I will nonetheless immediately tell you what your age is. How do I do this? Let’s do the algebra. 10X – 9Y = 10X – 10Y + Y = 10(X – Y) + Y Now remember, you asked an adult, so his/her age is a two digit number (maybe even 3 digits), thus reducing it by the single digit multiplied by nine is still positive – the lowest is can be is 100 – 81 which yields 19. Now make two numbers out of the result. The last digit and the number before it. This number is X – Y or the age minus the single digit you selected. The last digit is this very single digit. This is always so regardless of the digit you selected. So… Add tis digit to the other number and you get back the age! Q.E.D Example: I am 76 years old and here is what happens when I do the steps 76 x 10 = 760 760 – 18 = 742 made of 74 and 2. My age is 74 + 2 760 – 81 = 679 made of 67 and 9. My age is 67 + 9 A note to the socially aware mathemagician – it is safer to do it with a man. The chances of a veracious answer are much, much higher! The trick may be accomplished on any 2 or 3 digit number, not just one’s age, but if you want to know your date’s age, it’s a good way to elicit it. That’s All Folks PS for more Ageless “Age” mathemagics go to www.mgsltns.com/games.htm and also here: http://geekswithblogs.net/PointsToShare/archive/2011/11/15/mathemagics---guess-my-age---method-2.aspx

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