Search Results

Search found 235 results on 10 pages for 'l2'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 1/10 | 1 2 3 4 5 6 7 8 9 10  | Next Page >

  • Programmatically measure size and way-order of L1 and L2 caches

    - by osgx
    How can I measure programmatically (not query the OS, but measure) the size and order of associativity of L1 and L2 caches (data caches)? Assumptions about system: It has L1 and L2 cache (may be L3 too, may be cache sharing), It may have a hardware prefetch unit (just like P4+), It has a stable clocksource (tickcounter or good HPET for gettimeofday). There are no assumptions about OS (it can be Linux, Windows, or something non-standard), and we can't use POSIX queries. Language is C. And compiler optimizations may be disabled.

    Read the article

  • Programicaly measure size and way-order of L1 and L2 caches

    - by osgx
    Hello How can I measure programicaly (not query the OS, but measure) the size and order of associativity of L1 and L2 caches (data caches)? Assumtions about system: It has L1 and L2 cache (may be L3 too, may be cache sharing), It may have a hardware prefetch unit (just like P4+), it has a stable clocksource (tickcounter or good HPET for gettimeofday). There are no assumtions about OS (it can be Linux, Windows, smth non-standart), and we can't use posix queries. Language is C. And Compiler optimizations may be disabled.

    Read the article

  • NHibernate L2 Cache - fluent nHibernate configuration

    - by AWC
    I've managed to configure the L2 cache for Get\Load in FHN, but it's not working for queries configured using the ICriteria interface - it doesn't cache the results from these queries. Does anyone know why? The configurations are as follows: ICriteria: return unitOfWork .CurrentSession .CreateCriteria(typeof(Country)) .SetCacheable(true); Entity Mapping: public sealed class CountryMap : ClassMap<Country>, IMap { public CountryMap() { Table("Countries"); Not.LazyLoad(); Cache.ReadWrite().IncludeAll(); Id(x => x.Id); Map(x => x.TwoLetter); Map(x => x.ThreeLetter); Map(x => x.Name); } } And the session factory configuration for the database property: return () => MsSqlConfiguration.MsSql2005 .ConnectionString(BuildConnectionString()) .ShowSql() .Cache(c => c.UseQueryCache() .QueryCacheFactory<StandardQueryCacheFactory>() .ProviderClass(configuration.RepositoryCacheType) .UseMinimalPuts()) .FormatSql() .UseReflectionOptimizer(); Cheers AWC

    Read the article

  • Why there is no scoped locks for multiple mutexes in C++0x or Boost.Thread?

    - by Vicente Botet Escriba
    C++0x thread library or Boost.thread define non-member variadic template function that lock all lock avoiding dead lock. template <class L1, class L2, class... L3> void lock(L1&, L2&, L3&...); While this function avoid help to deadlock, the standard do not includes the associated scoped lock to write exception safe code. { std::lock(l1,l2); // do some thing // unlock li l2 exception safe } That means that we need to use other mechanism as try-catch block to make exception safe code or define our own scoped lock on multiple mutexes ourselves or even do that { std::lock(l1,l2); std::unique_lock lk1(l1, std::adopted); std::unique_lock lk2(l2, std::adopted); // do some thing // unlock li l2 on destruction of lk1 lk2 } Why the standard doesn't includes a scoped lock on multiple mutexes of the same type, as for example { std::array_unique_lock<std::mutex> lk(l1,l2); // do some thing // unlock l1 l2 on destruction of lk } or tuples of mutexes { std::tuple_unique_lock<std::mutex, std::recursive_mutex> lk(l1,l2); // do some thing // unlock l1 l2 on destruction of lk } Is there something wrong on the design?

    Read the article

  • Switch case assembly level code

    - by puffadder
    Hi All, I am programming C on cygwin windows. After having done a bit of C programming and getting comfortable with the language, I wanted to look under the hood and see what the compiler is doing for the code that I write. So I wrote down a code block containing switch case statements and converted them into assembly using: gcc -S foo.c Here is the C source: switch(i) { case 1: { printf("Case 1\n"); break; } case 2: { printf("Case 2\n"); break; } case 3: { printf("Case 3\n"); break; } case 4: { printf("Case 4\n"); break; } case 5: { printf("Case 5\n"); break; } case 6: { printf("Case 6\n"); break; } case 7: { printf("Case 7\n"); break; } case 8: { printf("Case 8\n"); break; } case 9: { printf("Case 9\n"); break; } case 10: { printf("Case 10\n"); break; } default: { printf("Nothing\n"); break; } } Now the resultant assembly for the same is: movl $5, -4(%ebp) cmpl $10, -4(%ebp) ja L13 movl -4(%ebp), %eax sall $2, %eax movl L14(%eax), %eax jmp *%eax .section .rdata,"dr" .align 4 L14: .long L13 .long L3 .long L4 .long L5 .long L6 .long L7 .long L8 .long L9 .long L10 .long L11 .long L12 .text L3: movl $LC0, (%esp) call _printf jmp L2 L4: movl $LC1, (%esp) call _printf jmp L2 L5: movl $LC2, (%esp) call _printf jmp L2 L6: movl $LC3, (%esp) call _printf jmp L2 L7: movl $LC4, (%esp) call _printf jmp L2 L8: movl $LC5, (%esp) call _printf jmp L2 L9: movl $LC6, (%esp) call _printf jmp L2 L10: movl $LC7, (%esp) call _printf jmp L2 L11: movl $LC8, (%esp) call _printf jmp L2 L12: movl $LC9, (%esp) call _printf jmp L2 L13: movl $LC10, (%esp) call _printf L2: Now, in the assembly, the code is first checking the last case (i.e. case 10) first. This is very strange. And then it is copying 'i' into 'eax' and doing things that are beyond me. I have heard that the compiler implements some jump table for switch..case. Is it what this code is doing? Or what is it doing and why? Because in case of less number of cases, the code is pretty similar to that generated for if...else ladder, but when number of cases increases, this unusual-looking implementation is seen. Thanks in advance.

    Read the article

  • tc u32 --- how to match L2 protocols in recent kernels?

    - by brownian
    I have a nice shaper, with hashed filtering, built at a linux bridge. In short, br0 connects external and internal physical interfaces, VLAN tagged packets are bridged "transparently" (I mean, no VLAN interfaces are there). Now, different kernels do it differently. I can be wrong with exact kernel verions ranges, please forgive me. Thanks. 2.6.26 So, in debian, 2.6.26 and up (up to 2.6.32, I believe) --- this works: tc filter add dev internal protocol 802.1q parent 1:0 prio 100 \ u32 ht 1:64 match ip dst 192.168.1.100 flowid 1:200 Here, "kernel" matches two bytes in "protocol" field with 0x8100, but counts the beginning of ip packet as a "zero position" (sorry for my English, if I'm a bit unclear). 2.6.32 Again, in debian (I've not built vanilla kernel), 2.6.32-5 --- this works: tc filter add dev internal protocol 802.1q parent 1:0 prio 100 \ u32 ht 1:64 match ip dst 192.168.1.100 at 20 flowid 1:200 Here, "kernel" matches the same for protocol, but counts offset from the beginning of this protocol's header --- I have to add 4 bytes to offset (20, not 16 for dst address). It's ok, seems more logical, as for me. 3.2.11, the latest stable now This works --- as if there is no 802.1q tag at all: tc filter add dev internal protocol ip parent 1:0 prio 100 \ u32 ht 1:64 match ip dst 192.168.1.100 flowid 1:200 The problem is that I couldn't find a way to match 802.1q tag so far. Matching 802.1q tag at past I could do this before as follows: tc filter add dev internal protocol 802.1q parent 1:0 prio 100 \ u32 match u16 0x0ed8 0x0fff at -4 flowid 1:300 Now I'm unable to match 802.1q tag with at 0, at -2, at -4, at -6 or like that. The main issue that I have zero hits count --- this filter is not being checked at all, "wrong protocol", in other words. Please, anyone, help me :-) Thanks!

    Read the article

  • Using recursion an append in prolog

    - by Adrian
    Lets say that I would like to construct a list (L2) by appending elements of another list (L) one by one. The result should be exactly the same as the input. This task is silly, but it'll help me understand how to recurse through a list and remove certain elements. I have put together the following code: create(L, L2) :- (\+ (L == []) -> L=[H|T], append([H], create(T, L2), L2);[]). calling it by create([1,2,3,4], L2) returns L2 = [1|create([2,3,4], **)\. which is not a desired result.

    Read the article

  • substitute in a nested list (prolog)

    - by linda
    /* substitute(X,Y,Xs,Ys) is true if the list Ys is the result of substituting Y for all occurrences of X in the list Xs. This is what I have so far: subs(_,_,[],[]). subs(X,Y,[X|L1],[Y|L2]):- subs(X,Y,L1,L2). subs(X,Y,[H|L1],[H|L2]):- X\=H, not(H=[_|_]), subs(X,Y,L1,L2). subs(X,Y,[H|_],[L2]):- X\=H, H=[_|_], subs(X,Y,H,L2). My code works except it omits the elements following the nested list. For example: ?- subs(a,b,[a,[a,c],a],Z). Z = [b, [b, c]] . What should I add to this program?

    Read the article

  • Using recustion an append in prolog

    - by Adrian
    Lets say that I would like to construct a list (L2) by appending elements of another list (L) one by one. The result should be exactly the same as the input. This task is silly, but it'll help me understand how to recurse through a list and remove certain elements. I have put together the following code: create(L, L2) :- (\+ (L == []) -> L=[H|T], append([H], create(T, L2), L2);[]). calling it by create([1,2,3,4], L2) returns L2 = [1|create([2,3,4], **)\. which is not a desired result.

    Read the article

  • [C++] My First Go With Function Templates

    - by bobber205
    Thought it was pretty straight forward. But I get a "iterator not dereferencable" errro when running the below code. What's wrong? template<typename T> struct SumsTo : public std::binary_function<T, T, bool> { int myInt; SumsTo(int a) { myInt = a; } bool operator()(const T& l, const T& r) { cout << l << " + " << r; if ((l + r) == myInt) { cout << " does add to " << myInt; } else { cout << " DOES NOT add to " << myInt; } return true; } }; void main() { list<int> l1; l1.push_back(1); l1.push_back(2); l1.push_back(3); l1.push_back(4); list<int> l2; l2.push_back(9); l2.push_back(8); l2.push_back(7); l2.push_back(6); transform(l1.begin(), l1.end(), l2.begin(), l2.end(), SumsTo<int>(10) ); }

    Read the article

  • Comparing lists in Standard ML

    - by user1050640
    I am extremely new to SML and we just got out first programming assignment for class and I need a little insight. The question is: write an ML function, called minus: int list * int list -> int list, that takes two non-decreasing integer lists and produces a non-decreasing integer list obtained by removing the elements from the first input list which are also found in the second input list. For example, minus( [1,1,1,2,2], [1,1,2,3] ) = [1,2] minus( [1,1,2,3],[1,1,1,2,2] ) = [3] Here is my attempt at answering the question. Can anyone tell me what I am doing incorrectly? I don't quite understand parsing lists. fun minus(xs,nil) = [] | minus(nil,ys) = [] | minus(x::xs,y::ys) = if x=y then minus(xs,ys) else x :: minus(x,ys); Here is a fix I just did, I think this is right now? fun minus(L1,nil) = L1 | minus(nil,L2) = [] | minus(L1,L2) = if hd(L1) > hd(L2) then minus(L1,tl(L2)) else if hd(L1) = hd(L2) then minus(tl(L1),tl(L2)) else hd(L1) :: minus(tl(L1), L2);

    Read the article

  • Finding 'free' times in MySQL

    - by James Inman
    Hi, I've got a table as follows: mysql> DESCRIBE student_lectures; +------------------+----------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------------+----------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | course_module_id | int(11) | YES | MUL | NULL | | | day | int(11) | YES | | NULL | | | start | datetime | YES | | NULL | | | end | datetime | YES | | NULL | | | cancelled_at | datetime | YES | | NULL | | | lecture_type_id | int(11) | YES | | NULL | | | lecture_id | int(11) | YES | | NULL | | | student_id | int(11) | YES | | NULL | | | created_at | datetime | YES | | NULL | | | updated_at | datetime | YES | | NULL | | +------------------+----------+------+-----+---------+----------------+ I'm essentially wanting to find times when a lecture doesn't happen - so to do this I'm thinking a query to group overlapping lectures together (so, for example, 9am-10am and 10am-11am lectures will be shown as a single 9am-11am lecture). There may be more than two lectures back-to-back. I've currently got this: SELECT l.start, l2.end FROM student_lectures l LEFT JOIN student_lectures l2 ON ( l2.start = l.end ) WHERE l.student_id = 1 AND l.start >= '2010-04-26 09:00:00' AND l.end <= '2010-04-30 19:00:00' AND l2.end IS NOT NULL AND l2.end != l.start GROUP BY l.start, l2.end ORDER BY l.start, l2.start Which returns: +---------------------+---------------------+ | start | end | +---------------------+---------------------+ | 2010-04-26 09:00:00 | 2010-04-26 11:00:00 | | 2010-04-26 10:00:00 | 2010-04-26 12:00:00 | | 2010-04-26 10:00:00 | 2010-04-26 13:00:00 | | 2010-04-26 13:15:00 | 2010-04-26 16:15:00 | | 2010-04-26 14:15:00 | 2010-04-26 16:15:00 | | 2010-04-26 15:15:00 | 2010-04-26 17:15:00 | | 2010-04-26 16:15:00 | 2010-04-26 18:15:00 | ...etc... The output I'm looking for from this would be: +---------------------+---------------------+ | start | end | +---------------------+---------------------+ | 2010-04-26 09:00:00 | 2010-04-26 13:00:00 | | 2010-04-26 13:15:00 | 2010-04-26 18:15:00 | Any help appreciated, thanks!

    Read the article

  • segmentation fault when using pointer to pointer

    - by user3697730
    I had been trying to use a pointer to pointer in a function,but is seems that I am not doing the memory allocation correctly... My code is: #include<stdio.h> #include<math.h> #include<ctype.h> #include<stdlib.h> #include<string.h> struct list{ int data; struct list *next; }; void abc (struct list **l,struct list **l2) { *l2=NULL; l2=(struct list**)malloc( sizeof(struct list*)); (*l)->data=12; printf("%d",(*l)->data); (*l2)->next=*l2; } int main() { struct list *l,*l2; abc(&l,&l2); system("pause"); return(0); } This code compiles,but I cannot run thw program..I get a segmentation fault..What should I do?Any help would be appreciated!

    Read the article

  • Java JPanel not showing up....

    - by user69514
    I'm not sure what I am doing wrong, but the text for my JPanels is not showing up. I just get the question number text, but the question is not showing up. Any ideas what I am doing wrong? import java.awt.*; import java.awt.event.*; import javax.swing.*; import javax.swing.event.*; class NewFrame extends JFrame { JPanel centerpanel; // For the questions. CardLayout card; // For the centerpanel. JTextField tf; // Used in question 1. boolean // Store selections for Q2. q2Option1, q2Option2, q2Option3, q2Option4; JList q4List; // For question 4. double // Score on each question. q1Score = 0, q2Score = 0, q3Score = 0, q4Score = 0; // Constructor. public NewFrame (int width, int height) { this.setTitle ("Snoot Club Membership Test"); this.setResizable (true); this.setSize (width, height); Container cPane = this.getContentPane(); // cPane.setLayout (new BorderLayout()); // First, a welcome message, as a Label. JLabel L = new JLabel ("<html><b>Are you elitist enough for our exclusive club?" + " <br>Fill out the form and find out</b></html>"); L.setForeground (Color.blue); cPane.add (L, BorderLayout.NORTH); // Now the center panel with the questions. card = new CardLayout (); centerpanel = new JPanel (); centerpanel.setLayout (card); centerpanel.setOpaque (false); // Each question will be created in a separate method. // The cardlayout requires a label as second parameter. centerpanel.add (firstQuestion (), "1"); centerpanel.add (secondQuestion(), "2"); centerpanel.add (thirdQuestion(), "3"); centerpanel.add (fourthQuestion(), "4"); cPane.add (centerpanel, BorderLayout.CENTER); // Next, a panel of four buttons at the bottom. // The four buttons: quit, submit, next-question, previous-question. JPanel bottomPanel = getBottomPanel (); cPane.add (bottomPanel, BorderLayout.SOUTH); // Finally, show the frame. this.setVisible (true); } // No-parameter constructor. public NewFrame () { this (500, 300); } // The first question uses labels for the question and // gets input via a textfield. A panel containing all // these things is returned. The question asks for // a vacation destination: the more exotic the location, // the higher the score. JPanel firstQuestion () { // We will package everything into a panel and return the panel. JPanel subpanel = new JPanel (); // We will place things in a single column, so // a GridLayout with one column is appropriate. subpanel.setLayout (new GridLayout (8,1)); JLabel L1 = new JLabel ("Question 1:"); L1.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L1); JLabel L2 = new JLabel (" Select a vacation destination"); L2.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L2); JLabel L3 = new JLabel (" 1. Baltimore"); L3.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L3); JLabel L4 = new JLabel (" 2. Disneyland"); L4.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L4); JLabel L5 = new JLabel (" 3. Grand Canyon"); L5.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L5); JLabel L6 = new JLabel (" 4. French Riviera"); L6.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L6); JLabel L7 = new JLabel ("Enter 1,2,3 or 4 below:"); L7.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L7); // Here's the textfield to get user-input. tf = new JTextField (); tf.addActionListener ( new ActionListener () { // This interface has only one method. public void actionPerformed (ActionEvent a) { String q1String = a.getActionCommand(); if (q1String.equals ("2")) q1Score = 2; else if (q1String.equals ("3")) q1Score = 3; else if (q1String.equals ("4")) q1Score = 4; else q1Score = 1; } } ); subpanel.add (tf); return subpanel; } // For the second question, a collection of checkboxes // will be used. More than one selection can be made. // A listener is required for each checkbox. The state // of each checkbox is recorded. JPanel secondQuestion () { JPanel subpanel = new JPanel (); subpanel.setLayout (new GridLayout (7,1)); JLabel L1 = new JLabel ("Question 2:"); L1.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L1); JLabel L2 = new JLabel (" Select ONE OR MORE things that "); L2.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L2); JLabel L3 = new JLabel (" you put into your lunch sandwich"); L3.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L3); // Initialize the selections to false. q2Option1 = q2Option2 = q2Option3 = q2Option4 = false; // First checkbox. JCheckBox c1 = new JCheckBox ("Ham, beef or turkey"); c1.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JCheckBox c = (JCheckBox) i.getSource(); q2Option1 = c.isSelected(); } } ); subpanel.add (c1); // Second checkbox. JCheckBox c2 = new JCheckBox ("Cheese"); c2.addItemListener ( new ItemListener () { // This is where we will react to a change in checkbox. public void itemStateChanged (ItemEvent i) { JCheckBox c = (JCheckBox) i.getSource(); q2Option2 = c.isSelected(); } } ); subpanel.add (c2); // Third checkbox. JCheckBox c3 = new JCheckBox ("Sun-dried Arugula leaves"); c3.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JCheckBox c = (JCheckBox) i.getSource(); q2Option3 = c.isSelected(); } } ); subpanel.add (c3); // Fourth checkbox. JCheckBox c4 = new JCheckBox ("Lemon-enhanced smoked Siberian caviar"); c4.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JCheckBox c = (JCheckBox) i.getSource(); q2Option4 = c.isSelected(); } } ); subpanel.add (c4); return subpanel; } // The third question allows only one among four choices // to be selected. We will use radio buttons. JPanel thirdQuestion () { JPanel subpanel = new JPanel (); subpanel.setLayout (new GridLayout (6,1)); JLabel L1 = new JLabel ("Question 3:"); L1.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L1); JLabel L2 = new JLabel (" And which mustard do you use?"); L2.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L2); // First, create the ButtonGroup instance. // We will add radio buttons to this group. ButtonGroup bGroup = new ButtonGroup(); // First checkbox. JRadioButton r1 = new JRadioButton ("Who cares?"); r1.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JRadioButton r = (JRadioButton) i.getSource(); if (r.isSelected()) q3Score = 1; } } ); bGroup.add (r1); subpanel.add (r1); // Second checkbox. JRadioButton r2 = new JRadioButton ("Safeway Brand"); r2.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JRadioButton r = (JRadioButton) i.getSource(); if (r.isSelected()) q3Score = 2; } } ); bGroup.add (r2); subpanel.add (r2); // Third checkbox. JRadioButton r3 = new JRadioButton ("Fleishman's"); r3.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JRadioButton r = (JRadioButton) i.getSource(); if (r.isSelected()) q3Score = 3; } } ); bGroup.add (r3); subpanel.add (r3); // Fourth checkbox. JRadioButton r4 = new JRadioButton ("Grey Poupon"); r4.addItemListener ( new ItemListener () { public void itemStateChanged (ItemEvent i) { JRadioButton r = (JRadioButton) i.getSource(); if (r.isSelected()) q3Score = 4; } } ); bGroup.add (r4); subpanel.add (r4); return subpanel; } // For the fourth question we will use a drop-down Choice. JPanel fourthQuestion () { JPanel subpanel = new JPanel (); subpanel.setLayout (new GridLayout (3,1)); JLabel L1 = new JLabel ("Question 4:"); L1.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L1); JLabel L2 = new JLabel (" Your movie preference, among these:"); L2.setFont (new Font ("SansSerif", Font.ITALIC, 15)); subpanel.add (L2); // Create a JList with options. String[] movies = { "Lethal Weapon IV", "Titanic", "Saving Private Ryan", "Le Art Movie avec subtitles"}; q4List = new JList (movies); q4Score = 1; q4List.addListSelectionListener ( new ListSelectionListener () { public void valueChanged (ListSelectionEvent e) { q4Score = 1 + q4List.getSelectedIndex(); } } ); subpanel.add (q4List); return subpanel; } void computeResult () { // Clear the center panel. centerpanel.removeAll(); // Create a new panel to display in the center. JPanel subpanel = new JPanel (new GridLayout (5,1)); // Score on question 1. JLabel L1 = new JLabel ("Score on question 1: " + q1Score); L1.setFont (new Font ("Serif", Font.ITALIC, 15)); subpanel.add (L1); // Score on question 2. if (q2Option1) q2Score += 1; if (q2Option2) q2Score += 2; if (q2Option3) q2Score += 3; if (q2Option4) q2Score += 4; q2Score = 0.6 * q2Score; JLabel L2 = new JLabel ("Score on question 2: " + q2Score); L2.setFont (new Font ("Serif", Font.ITALIC, 15)); subpanel.add (L2); // Score on question 3. JLabel L3 = new JLabel ("Score on question 3: " + q3Score); L3.setFont (new Font ("Serif", Font.ITALIC, 15)); subpanel.add (L3); // Score on question 4. JLabel L4 = new JLabel ("Score on question 4: " + q4Score); L4.setFont (new Font ("Serif", Font.ITALIC, 15)); subpanel.add (L4); // Weighted score. double avg = (q1Score + q2Score + q3Score + q4Score) / (double) 4; JLabel L5; if (avg <= 3.5) L5 = new JLabel ("Your average score: " + avg + " - REJECTED!"); else L5 = new JLabel ("Your average score: " + avg + " - WELCOME!"); L5.setFont (new Font ("Serif", Font.BOLD, 20)); //L5.setAlignment (JLabel.CENTER); subpanel.add (L5); // Now add the new subpanel. centerpanel.add (subpanel, "5"); // Need to mark the centerpanel as "altered" centerpanel.invalidate(); // Everything "invalid" (e.g., the centerpanel above) // is now re-computed. this.validate(); } JPanel getBottomPanel () { // Create a panel into which we will place buttons. JPanel bottomPanel = new JPanel (); // A "previous-question" button. JButton backward = new JButton ("Previous question"); backward.setFont (new Font ("Serif", Font.PLAIN | Font.BOLD, 15)); backward.addActionListener ( new ActionListener () { public void actionPerformed (ActionEvent a) { // Go back in the card layout. card.previous (centerpanel); } } ); bottomPanel.add (backward); // A forward button. JButton forward = new JButton ("Next question"); forward.setFont (new Font ("Serif", Font.PLAIN | Font.BOLD, 15)); forward.addActionListener ( new ActionListener () { public void actionPerformed (ActionEvent a) { // Go forward in the card layout. card.next (centerpanel); } } ); bottomPanel.add (forward); // A submit button. JButton submit = new JButton ("Submit"); submit.setFont (new Font ("Serif", Font.PLAIN | Font.BOLD, 15)); submit.addActionListener ( new ActionListener () { public void actionPerformed (ActionEvent a) { // Perform submit task. computeResult(); } } ); bottomPanel.add (submit); JButton quitb = new JButton ("Quit"); quitb.setFont (new Font ("Serif", Font.PLAIN | Font.BOLD, 15)); quitb.addActionListener ( new ActionListener () { public void actionPerformed (ActionEvent a) { System.exit (0); } } ); bottomPanel.add (quitb); return bottomPanel; } } public class Survey { public static void main (String[] argv) { NewFrame nf = new NewFrame (600, 300); } }

    Read the article

  • TcpListener problem - re-binding to same port with different local addresses

    - by Zvika
    I'm trying to do the following: listen on some port for loopback connections only, and then start listening on any IP address. Here is the code: TcpListener l1 = new TcpListener(new IPEndPoint(IPAddress.Loopback, 12345)); l1.Start(); Socket s = l1.AcceptSocket(); Console.ReadKey(); //s.Close(); l1.Stop(); TcpListener l2 = new TcpListener(new IPEndPoint(IPAddress.Any, 12345)); l2.Start(); l2.AcceptSocket(); Console.ReadKey(); The problem is that if a client connects while listening on the Loopback address (l1), then no other client can connect to the Loopback address when the second listener (l2) starts listening. why is that? Another thing I noticed is that if I close all clients that connected to l1 (the remarked line), then l2 does accept loopback connections. Any ideas?

    Read the article

  • Generate Permutations of a List

    - by Eric Mercer
    I'm writing a function that takes a list and returns a list of permutations of the argument. I know how to do it by using a function that removes an element and then recursively use that function to generate all permutations. I now have a problem where I want to use the following function: (define (insert-everywhere item lst) (define (helper item L1 L2) (if (null? L2) (cons (append L1 (cons item '())) '()) (cons (append L1 (cons item L2)) (helper item (append L1 (cons (car L2) '())) (cdr L2))))) (helper item '() lst)) This function will insert the item into every possible location of the list, like the following: (insert-everywhere 1 '(a b)) will get: '((1 a b) (a 1 b) (a b 1)) How would I use this function to get all permutations of a list? I now have: (define (permutations lst) (if (null? lst) '() (insert-helper (car lst) (permutations (cdr lst))))) (define (insert-helper item lst) (cond ((null? lst) '()) (else (append (insert-everywhere item (car lst)) (insert-helper item (cdr lst)))))) but doing (permutations '(1 2 3)) just returns the empty list '().

    Read the article

  • list within a list

    - by atm atm
    I'm working on this problem, but I cannot figure out the second part. I tried using reverse list but it did not work out how I planned it. Given a list L (e.g. [1,2,3,4]), write a program that generates the following nested lists: L1 = [[1],[1,2],[1,2,3],[1,2,3,4]], L2 = [[4],[3,4],[2,3,4],[1,2,3,4]]. My code that I have so far: mylist=[,1,2,3,4] print("Orginal list L=",mylist) n=len(mylist) l1=[] l2=[] for x in range(1,n+1,1): l1.append(mylist[0:x]) print("L1=",l1) #prints final product of l1 mylist.reverse() #this is where i get messed up for x in range(1,n+1,1): l2.append(mylist[0:x]) print("L2=",l2)

    Read the article

  • T4 Performance Counters explained

    - by user13346607
    Now that T4 is out for a few month some people might have wondered what details of the new pipeline you can monitor. A "cpustat -h" lists a lot of events that can be monitored, and only very few are self-explanatory. I will try to give some insight on all of them, some of these "PIC events" require an in-depth knowledge of T4 pipeline. Over time I will try to explain these, for the time being these events should simply be ignored. (Side note: some counters changed from tape-out 1.1 (*only* used in the T4 beta program) to tape-out 1.2 (used in the systems shipping today) The table only lists the tape-out 1.2 counters) 0 0 1 1058 6033 Oracle Microelectronics 50 14 7077 14.0 Normal 0 false false false EN-US JA X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Cambria; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} pic name (cpustat) Prose Comment Sel-pipe-drain-cycles, Sel-0-[wait|ready], Sel-[1,2] Sel-0-wait counts cycles a strand waits to be selected. Some reasons can be counted in detail; these are: Sel-0-ready: Cycles a strand was ready but not selected, that can signal pipeline oversubscription Sel-1: Cycles only one instruction or µop was selected Sel-2: Cycles two instructions or µops were selected Sel-pipe-drain-cycles: cf. PRM footnote 8 to table 10.2 Pick-any, Pick-[0|1|2|3] Cycles one, two, three, no or at least one instruction or µop is picked Instr_FGU_crypto Number of FGU or crypto instructions executed on that vcpu Instr_ld dto. for load Instr_st dto. for store SPR_ring_ops dto. for SPR ring ops Instr_other dto. for all other instructions not listed above, PRM footnote 7 to table 10.2 lists the instructions Instr_all total number of instructions executed on that vcpu Sw_count_intr Nr of S/W count instructions on that vcpu (sethi %hi(fc000),%g0 (whatever that is))  Atomics nr of atomic ops, which are LDSTUB/a, CASA/XA, and SWAP/A SW_prefetch Nr of PREFETCH or PREFETCHA instructions Block_ld_st Block loads or store on that vcpu IC_miss_nospec, IC_miss_[L2_or_L3|local|remote]\ _hit_nospec Various I$ misses, distinguished by where they hit. All of these count per thread, but only primary events: T4 counts only the first occurence of an I$ miss on a core for a certain instruction. If one strand misses in I$ this miss is counted, but if a second strand on the same core misses while the first miss is being resolved, that second miss is not counted This flavour of I$ misses counts only misses that are caused by instruction that really commit (note the "_nospec") BTC_miss Branch target cache miss ITLB_miss ITLB misses (synchronously counted) ITLB_miss_asynch dto. but asynchronously [I|D]TLB_fill_\ [8KB|64KB|4MB|256MB|2GB|trap] H/W tablewalk events that fill ITLB or DTLB with translation for the corresponding page size. The “_trap” event occurs if the HWTW was not able to fill the corresponding TLB IC_mtag_miss, IC_mtag_miss_\ [ptag_hit|ptag_miss|\ ptag_hit_way_mismatch] I$ micro tag misses, with some options for drill down Fetch-0, Fetch-0-all fetch-0 counts nr of cycles nothing was fetched for this particular strand, fetch-0-all counts cycles nothing was fetched for all strands on a core Instr_buffer_full Cycles the instruction buffer for a strand was full, thereby preventing any fetch BTC_targ_incorrect Counts all occurences of wrongly predicted branch targets from the BTC [PQ|ROB|LB|ROB_LB|SB|\ ROB_SB|LB_SB|RB_LB_SB|\ DTLB_miss]\ _tag_wait ST_q_tag_wait is listed under sl=20. These counters monitor pipeline behaviour therefore they are not strand specific: PQ_...: cycles Rename stage waits for a Pick Queue tag (might signal memory bound workload for single thread mode, cf. Mail from Richard Smith) ROB_...: cycles Select stage waits for a ROB (ReOrderBuffer) tag LB_...: cycles Select stage waits for a Load Buffer tag SB_...: cycles Select stage waits for Store Buffer tag combinations of the above are allowed, although some of these events can overlap, the counter will only be incremented once per cycle if any of these occur DTLB_...: cycles load or store instructions wait at Pick stage for a DTLB miss tag [ID]TLB_HWTW_\ [L2_hit|L3_hit|L3_miss|all] Counters for HWTW accesses caused by either DTLB or ITLB misses. Canbe further detailed by where they hit IC_miss_L2_L3_hit, IC_miss_local_remote_remL3_hit, IC_miss I$ prefetches that were dropped because they either miss in L2$ or L3$ This variant counts misses regardless if the causing instruction commits or not DC_miss_nospec, DC_miss_[L2_L3|local|remote_L3]\ _hit_nospec D$ misses either in general or detailed by where they hit cf. the explanation for the IC_miss in two flavours for an explanation of _nospec and the reasoning for two DC_miss counters DTLB_miss_asynch counts all DTLB misses asynchronously, there is no way to count them synchronously DC_pref_drop_DC_hit, SW_pref_drop_[DC_hit|buffer_full] L1-D$ h/w prefetches that were dropped because of a D$ hit, counted per core. The others count software prefetches per strand [Full|Partial]_RAW_hit_st_[buf|q] Count events where a load wants to get data that has not yet been stored, i. e. it is still inside the pipeline. The data might be either still in the store buffer or in the store queue. If the load's data matches in the SB and in the store queue the data in buffer takes precedence of course since it is younger [IC|DC]_evict_invalid, [IC|DC|L1]_snoop_invalid, [IC|DC|L1]_invalid_all Counter for invalidated cache evictions per core St_q_tag_wait Number of cycles pipeline waits for a store queue tag, of course counted per core Data_pref_[drop_L2|drop_L3|\ hit_L2|hit_L3|\ hit_local|hit_remote] Data prefetches that can be further detailed by either why they were dropped or where they did hit St_hit_[L2|L3], St_L2_[local|remote]_C2C, St_local, St_remote Store events distinguished by where they hit or where they cause a L2 cache-to-cache transfer, i.e. either a transfer from another L2$ on the same die or from a different die DC_miss, DC_miss_\ [L2_L3|local|remote]_hit D$ misses either in general or detailed by where they hit cf. the explanation for the IC_miss in two flavours for an explanation of _nospec and the reasoning for two DC_miss counters L2_[clean|dirty]_evict Per core clean or dirty L2$ evictions L2_fill_buf_full, L2_wb_buf_full, L2_miss_buf_full Per core L2$ buffer events, all count number of cycles that this state was present L2_pipe_stall Per core cycles pipeline stalled because of L2$ Branches Count branches (Tcc, DONE, RETRY, and SIT are not counted as branches) Br_taken Counts taken branches (Tcc, DONE, RETRY, and SIT are not counted as branches) Br_mispred, Br_dir_mispred, Br_trg_mispred, Br_trg_mispred_\ [far_tbl|indir_tbl|ret_stk] Counter for various branch misprediction events.  Cycles_user counts cycles, attribute setting hpriv, nouser, sys controls addess space to count in Commit-[0|1|2], Commit-0-all, Commit-1-or-2 Number of times either no, one, or two µops commit for a strand. Commit-0-all counts number of times no µop commits for the whole core, cf. footnote 11 to table 10.2 in PRM for a more detailed explanation on how this counters interacts with the privilege levels

    Read the article

  • Tail-recursive merge sort in OCaml

    - by CFP
    Hello world! I’m trying to implement a tail-recursive list-sorting function in OCaml, and I’ve come up with the following code: let tailrec_merge_sort l = let split l = let rec _split source left right = match source with | [] -> (left, right) | head :: tail -> _split tail right (head :: left) in _split l [] [] in let merge l1 l2 = let rec _merge l1 l2 result = match l1, l2 with | [], [] -> result | [], h :: t | h :: t, [] -> _merge [] t (h :: result) | h1 :: t1, h2 :: t2 -> if h1 < h2 then _merge t1 l2 (h1 :: result) else _merge l1 t2 (h2 :: result) in List.rev (_merge l1 l2 []) in let rec sort = function | [] -> [] | [a] -> [a] | list -> let left, right = split list in merge (sort left) (sort right) in sort l ;; Yet it seems that it is not actually tail-recursive, since I encounter a "Stack overflow during evaluation (looping recursion?)" error. Could you please help me spot the non tail-recursive call in this code? I've searched quite a lot, without finding it. Cout it be the let binding in the sort function? Thanks a lot, CFP.

    Read the article

  • custom C++ boost::lambda expression help

    - by aaa
    hello. A little bit of background: I have some strange multiple nested loops which I converted to flat work queue (basically collapse single index loops to single multi-index loop). right now each loop is hand coded. I am trying to generalized approach to work with any bounds using lambda expressions: For example: // RANGE(i,I,N) is basically a macro to generate `int i = I; i < N; ++i ` // for (RANGE(lb, N)) { // for (RANGE(jb, N)) { // for (RANGE(kb, max(lb, jb), N)) { // for (RANGE(ib, jb, kb+1)) { // is equivalent to something like (overload , to produce range) flat<1, 3, 2, 4>((_2, _3+1), (max(_4,_3), N), N, N) the internals of flat are something like: template<size_t I1, size_t I2, ..., class L1_, class L2, ..._> boost::array<int,4> flat(L1_ L1, L2_ L2, ...){ //boost::array<int,4> current; class variable bool advance; L2_ l2 = L2.bind(current); // bind current value to lambda { L1_ l1 = L1.bind(current); //bind current value to innermost lambda l1.next(); advance = !(l1 < l1.upper()); // some internal logic if (advance) { l2.next(); current[0] = l1.lower(); } } //..., } my question is, can you give me some ideas how to write lambda (derived from boost) which can be bound to index array reference to return upper, lower bounds according to lambda expression? thank you much

    Read the article

  • Java assert nasty side-effect - compiler bug?

    - by Alex
    This public class test { public static void main(String[] args) { Object o = null; assert o != null; if(o != null) System.out.println("o != null"); } } prints out "o != null"; both 1.5_22 and 1.6_18. Compiler bug? Commenting out the assert fixes it. The byte code appears to jump directly to the print statement when assertions are disabled: public static main(String[]) : void L0 LINENUMBER 5 L0 ACONST_NULL ASTORE 1 L1 LINENUMBER 6 L1 GETSTATIC test.$assertionsDisabled : boolean IFNE L2 ALOAD 1: o IFNONNULL L2 NEW AssertionError DUP INVOKESPECIAL AssertionError.<init>() : void ATHROW L2 LINENUMBER 8 L2 GETSTATIC System.out : PrintStream LDC "o != null" INVOKEVIRTUAL PrintStream.println(String) : void L3 LINENUMBER 9 L3 RETURN L4

    Read the article

  • Problem with Closure properties of context free languages

    - by altius
    hello i have the following sentence a language L1={a^n * b^n : n=0} and L2={b^n * a^n : n=0} are context free languages so they are close a=under the L1L2 so L={a^n * b^2n A^n : n=0} must be context free too because it is generated by a close property I have to prove if this sentence is true or not so i check the L language and i do not think that it is context free then i also saw that L2 is L1 reversed do i have to check if L1, L2 are deterministic ? please help because i am in a dead end

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

1 2 3 4 5 6 7 8 9 10  | Next Page >