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  • Having a number in

    - by Wiika
    Can someone give me a query that will return as a result rows ID 1 & 3? ID Name Hidden 1 Mika 1,4,2 2 Loca 0 3 Nosta 4 4 Like 2 Something like this SELECT * FROM table WHERE Hidden HAVING(4)

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  • Python calling class methods with the wrong number of parameters

    - by Hussain
    I'm just beginning to learn python. I wrote an example script to test OOP in python, but something very odd has happened. When I call a class method, Python is calling the function with one more parameter than given. Here is the code: 1. class Bar: 2. num1,num2 = 0,0 3. def __init__(num1,num2): 4. num1,num2 = num1,num2 5. def foo(): 6. if num1 num2: 7. print num1,'is greater than ',num2,'!' 8. elif num1 is num2: 9. print num1,' is equal to ',num2,'!' 10. else: 11. print num1,' is less than ',num2,'!' 12. a,b,t = 42,84,bar(a,b) 13. t.foo 14. 15. t.num1 = t.num1^t.num2 16. t.num2 = t.num2^t.num1 17. t.num1 = t.num1^t.num2 18. 19. t.foo 20. And the error message I get: python test.py Traceback (most recent call last): File "test.py", line 12, in a,b,t = 42,84,bar(a,b) NameError: name 'bar' is not defined Can anyone help? Thanks in advance

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  • Phone Number Recognition in Javascript

    - by samer
    Hi Guys, Is there a javascript library the can recognize phone numbers in a web page? Just like what skype did on their firefox plugin. Or do you know a way on how to do it? Websites or any tutorial that do the same would be very helpful. Your reply is greatly appreciated. Best,

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  • Internet Explorer not incrementing number for non-sibling <li> elements

    - by biagidp
    I've got some html that looks like this: <ol> <div> <li>one</li> </div> <div> <li>two</li> </div> <div> <li>three</li> </div> </ol> Which looks like this in Chrome/Firefox: 1. one 2. two 3. three But looks like this in IE: 1. one 1. two 1. three If I change the code so that the li element is the parent of the div element instead of the other way around (so that all the li elements are siblings) IE renders it correctly. Anyone know what causes this or if this is the intended working behavior of IE? Furthermore is one way technically more correct than the other? <div><li></li></div> VS. <li><div></div></li> Thanks, David

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  • Convert month number to month short name

    - by Roland
    I have a variable with the following value $month = 201002; the first 4 numbers represent the year, and the last 2 numbers represent the month. I need to get the last 2 numbers in the month string name eg. Feb My code looks like this <?php echo date('M',substr($month,4,6)); ?> I can I go about to obtain the month name

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  • adb doesn't get phone's device name/number

    - by Dona Hertel
    Okay, I have a strange problem I haven't seen listed anywhere. I'm developing an android app and I would like to run it on my Huawei Ascend. I have set up a file in /etc/udev/90-android.rules with the line: SUBSYSTEM=="usb", SYSFS{idVendor}=="12d1", MODE="0666" where '12d1' is the correct vendor ID for this phone (I verified this with 'lsusb' command). When I plug in the phone (it does have debugging on) and restart the adb server I get a connection but the name field does not get set. The output to 'adb devices' is: List of devices attached \n ???????????? device Plugging and unplugging the cable doesn't resolve this. Neither does restarting the adb server. Nor does a total reboot of both my computer or the phone. This is fine as I can get logs and a shell. The problem is that in the eclipse plugin, the device's name is list as "????????????" and so when it tries connect, it quits with an error message of 'device not found' even though the device is listed and 'online'. Is there something else I need to do? Do I need to set the name of the device somehow? cocofan P.S.: The app has 'debuggable' set to true in the manifest file.

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  • sorting, average and finding the lowest number from a static array Java

    - by user3701322
    i'm trying to input students and input their results for course work and exams and what i'm having trouble with is finding the average total score, the lowest total score and printing all students in order of total scores highest - lowest import java.util.*; import java.text.*; public class Results { static String[] name = new String[100]; static int[] coursework = new int[100]; static int[] exam = new int[100]; static int count = 0; public static void main(String[] args) { Scanner input = new Scanner(System.in); boolean flag = true; while(flag) { System.out.println( "1. Add Student\n" + "2. List All Students\n" + "3. List Student Grades\n" + "4. Total Score Average\n" + "5. Highest Total Score\n" + "6. Lowest Total Score\n" + "7. List all Students and Total Scores\n" + "8. Quit\n"); System.out.print("Enter choice (1 - 8): "); int choice = input.nextInt(); switch(choice) { case 1: add(); break; case 2: listAll(); break; case 3: listGrades(); break; case 4: average(); break; case 5: highestTotal(); break; case 6: lowestTotal(); break; case 7: order(); break; case 8: flag = false; break; default: System.out.println("\nNot an option\n"); } DateFormat dateFormat = new SimpleDateFormat("dd/MM/yyyy HH:mm:ss"); Date date = new Date(); System.out.println(dateFormat.format(date)); } System.out.println("\n\nHave a nice day"); }//end of main static void add() { Scanner input = new Scanner(System.in); System.out.println("Insert Name: "); String names = input.nextLine(); System.out.println("Insert Coursework: "); int courseworks = input.nextInt(); System.out.println("Insert Exam: "); int exams = input.nextInt(); name[count] = names; coursework[count] = courseworks; exam[count] = exams; count++; } static void listAll() { for(int i=0;i<count;i++) { System.out.printf("%s %d %d\n", name[i], coursework[i], exam[i]); } } static void listGrades() { for(int i=0;i<count;i++){ if(coursework[i] + exam[i] > 79) { System.out.println(name[i] + " HD"); } else if(coursework[i] + exam[i] > 69) { System.out.println(name[i] + " DI"); } else if(coursework[i] + exam[i] > 59) { System.out.println(name[i] + " CR"); } else if(coursework[i] + exam[i] > 49) { System.out.println(name[i] + " PA"); } else { System.out.println(name[i] + " NN"); } } } static void average() { } static void highestTotal() { int largest=exam[0]; String student=name[0]; for(int i=0; i<exam.length; i++){ if(exam[i]>largest){ largest = exam[i] + coursework[i]; student = name[i]; } } System.out.printf(student + ": "+ largest + "\n" ); } static void lowestTotal() { int min = 0; for(int i=0; i<=exam[i]; i++){ for(int j =0; j<=exam[i]; j++){ if(exam[i]<=exam[j] && j==exam[j]){ min = exam[i] + coursework[i]; } else{ continue; } } } System.out.printf(name + ": "+ min + "\n" ); } static void order() { } }

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  • java number exceeds long.max_value - how to detect?

    - by jurchiks
    I'm having problems detecting if a sum/multiplication of two numbers exceeds the maximum value of a long integer. Example code: long a = 2 * Long.MAX_VALUE; System.out.println("long.max * smth > long.max... or is it? a=" + a); This gives me -2, while I would expect it to throw a NumberFormatException... Is there a simple way of making this work? Because I have some code that does multiplications in nested IF blocks or additions in a loop and I would hate to add more IFs to each IF or inside the loop. Edit: oh well, it seems that this answer from another question is the most appropriate for what I need: http://stackoverflow.com/a/9057367/540394 I don't want to do boxing/unboxing as it adds unnecassary overhead, and this way is very short, which is a huge plus to me. I'll just write two short functions to do these checks and return the min or max long. Edit2: here's the function for limiting a long to its min/max value according to the answer I linked to above: /** * @param a : one of the two numbers added/multiplied * @param b : the other of the two numbers * @param c : the result of the addition/multiplication * @return the minimum or maximum value of a long integer if addition/multiplication of a and b is less than Long.MIN_VALUE or more than Long.MAX_VALUE */ public static long limitLong(long a, long b, long c) { return (((a > 0) && (b > 0) && (c <= 0)) ? Long.MAX_VALUE : (((a < 0) && (b < 0) && (c >= 0)) ? Long.MIN_VALUE : c)); } Tell me if you think this is wrong.

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  • Fast way to manually mod a number

    - by Nikolai Mushegian
    I need to be able to calculate (a^b) % c for very large values of a and b (which individually are pushing limit and which cause overflow errors when you try to calculate a^b). For small enough numbers, using the identity (a^b)%c = (a%c)^b%c works, but if c is too large this doesn't really help. I wrote a loop to do the mod operation manually, one a at a time: private static long no_Overflow_Mod(ulong num_base, ulong num_exponent, ulong mod) { long answer = 1; for (int x = 0; x < num_exponent; x++) { answer = (answer * num_base) % mod; } return answer; } but this takes a very long time. Is there any simple and fast way to do this operation without actually having to take a to the power of b AND without using time-consuming loops? If all else fails, I can make a bool array to represent a huge data type and figure out how to do this with bitwise operators, but there has to be a better way.

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  • Parallel Haskell in order to find the divisors of a huge number

    - by Dragno
    I have written the following program using Parallel Haskell to find the divisors of 1 billion. import Control.Parallel parfindDivisors :: Integer->[Integer] parfindDivisors n = f1 `par` (f2 `par` (f1 ++ f2)) where f1=filter g [1..(quot n 4)] f2=filter g [(quot n 4)+1..(quot n 2)] g z = n `rem` z == 0 main = print (parfindDivisors 1000000000) I've compiled the program with ghc -rtsopts -threaded findDivisors.hs and I run it with: findDivisors.exe +RTS -s -N2 -RTS I have found a 50% speedup compared to the simple version which is this: findDivisors :: Integer->[Integer] findDivisors n = filter g [1..(quot n 2)] where g z = n `rem` z == 0 My processor is a dual core 2 duo from Intel. I was wondering if there can be any improvement in above code. Because in the statistics that program prints says: Parallel GC work balance: 1.01 (16940708 / 16772868, ideal 2) and SPARKS: 2 (1 converted, 0 overflowed, 0 dud, 0 GC'd, 1 fizzled) What are these converted , overflowed , dud, GC'd, fizzled and how can help to improve the time.

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  • Minimum OS version number, iPhone app

    - by Michael Frost
    Hi all I've built an iPhone app which is live in the app-store. When originally submitting the app it showed up in App Store as requiring iPhone OS 3.1.3. When later updating the app I made sure my settings in Xcode for the target for the app store build had the Base SDK version set to 3.1.3 and the Deployment Target version set to 3.0, however it still shows up in app store as requiring 3.1.3. From what I've understood the Deployment Target version is the one setting the requirement in app store? Or is there any information concerning this that I should have updated in iTunes Connect when submitting the updated app? Thanks, Michael

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  • Generate a sequence of Fibonacci number in Scala

    - by qin
    def fibSeq(n: Int): List[Int] = { var ret = scala.collection.mutable.ListBuffer[Int](1, 2) while (ret(ret.length - 1) < n) { val temp = ret(ret.length - 1) + ret(ret.length - 2) if (temp >= n) { return ret.toList } ret += temp } ret.toList } So the above is my code to generate a Fibonacci sequence using Scala to a value n. I am wondering if there is a more elegant way to do this in Scala?

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  • Import a number of csv files and replace NA by Zeros

    - by tao.hong
    I know how to do this individually. However, I have more than 1000 files. I decided to use a for loop. However, it seems like I did not find the correct way to evaluate my variables. Here is my code setwd('C:/data') filenames=dir() #find file names for (i in filenames){ adt = substr(x = i, start = 1, stop = nchar(i)-4) name=paste("data_", adt, sep="") assign(name, read.csv(i,header=T,sep=",")) #read each file and assign a variable name starting with data_ to it func=paste('name[is.na(name)] <- 0',sep="") # here is the place I have problem. R will not consider name is a parameter whose values change in each iteration eval((text=func)) }

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  • Why is my number being rounded incorrectly?

    - by izb
    This feels like the kind of code that only fails in-situ, but I will attempt to adapt it into a code snippet that represents what I'm seeing. float f = myFloat * myConstInt; /* Where myFloat==13.45, and myConstInt==20 */ int i = (int)f; int i2 = (int)(myFloat * myConstInt); After stepping through the code, i==269, and i2==268. What's going on here to account for the difference?

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  • limit number of characters entered in textarea

    - by Abu Hamzah
    here is the script does what i want but not exactly, my question is, how can i stop user entering text once it reached the lmit of 255 characters? var limit = 255; var txt = $('textarea[id$=txtPurpose]'); $(txt).keyup(function() { var len = $(this).val().length; if (len > limit) { //this.value = this.value.substring(0, 50); $(this).addClass('goRed'); $('#spn').text(len - limit + " characters exceeded"); return false; } else { $(this).removeClass('goRed'); $('#spn').text(limit - len + " characters left"); } }); if there is a better way please let me know.

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  • What version number should an unreleased project receive?

    - by Byran
    Note: I'm new to version numbering. Please excuse my ignorance. I have a project where an attempted major release (Version B) was abandoned then later re-attempted and release (Version C). Each version has major changes from the previous version that I wouldn't consider an minor update. Little to nothing of Version B made it into Version C. Version A (1.0) Developed, released, updated, etc. Version B (???) Developed, suspended, abandoned. Version C (2.0) Developed, released, updated, etc. I feel like I should have version them like so, but worried about confusion of the missing version: Version A (1.0) Version B (2.0) Version C (3.0)

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