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  • Zero division does not throw exception in nunit

    - by Boris
    Running the following C# code through NUnit yields Test.ControllerTest.TestSanity: Expected: <System.DivideByZeroException> But was: null So either no DivideByZeroException is thrown, or NUnit does not catch it. Similar to this question, but the answers he got, do not seem to work for me. This is using NUnit 2.5.5.10112, and .NET 4.0.30319. [Test] public void TestSanity() { Assert.Throws<DivideByZeroException>(new TestDelegate(() => DivideByZero())); } private void DivideByZero() { // Parse "0" to make sure to get an error at run time, not compile time. var a = (1 / Double.Parse("0")); } Any ideas?

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  • A Review of From Zero To SSIS Training

    - by andyleonard
    I recently (5-9 Mar 2012) delivered my five-day SSIS training course – From Zero To SSIS! – in London. The class was delivered in collaboration with TechniTrain . I must commend Chris Webb ( blog | @Technitrain ) and Helen Lau on their leadership, professionalism, and attention to detail. They made the course a breeze for the students and the instructor! It was a pleasure and privilege to work with them. In addition to people just learning data integration, this class contained several experienced...(read more)

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  • trying to divide complex numbers, division by zero

    - by user553619
    I'm trying the program below to divide complex numbers, it works for complex numbers but not when the denominator is real (i.e, the complex part is zero). Division by zero occurs in this line ratio = b->r / b->i ;, when the complex part b->i is zero (in the case of a real denominator). How do I get around this? and why did the programmer do this, instead of the more straightforward rule for complex division The wikipedia rule seems to be better, and no division by zero error would occur here. Did I miss something? Why did the programmer not use the wikipedia formula?? Thanks /*! @file dcomplex.c * \brief Common arithmetic for complex type * * <pre> * -- SuperLU routine (version 2.0) -- * Univ. of California Berkeley, Xerox Palo Alto Research Center, * and Lawrence Berkeley National Lab. * November 15, 1997 * * This file defines common arithmetic operations for complex type. * </pre> */ #include <math.h> #include <stdlib.h> #include <stdio.h> #include "slu_dcomplex.h" /*! \brief Complex Division c = a/b */ void z_div(doublecomplex *c, doublecomplex *a, doublecomplex *b) { double ratio, den; double abr, abi, cr, ci; if( (abr = b->r) < 0.) abr = - abr; if( (abi = b->i) < 0.) abi = - abi; if( abr <= abi ) { if (abi == 0) { fprintf(stderr, "z_div.c: division by zero\n"); exit(-1); } ratio = b->r / b->i ; den = b->i * (1 + ratio*ratio); cr = (a->r*ratio + a->i) / den; ci = (a->i*ratio - a->r) / den; } else { ratio = b->i / b->r ; den = b->r * (1 + ratio*ratio); cr = (a->r + a->i*ratio) / den; ci = (a->i - a->r*ratio) / den; } c->r = cr; c->i = ci; }

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  • zero-config CGI enabled web server

    - by halp
    To serve static content of a directory over http, one can simply navigate to that directory and type: python -m SimpleHTTPServer 11111 which will start a http server on port 11111. This hack is nice because it requires zero-config: no stand-alone web server, no config files at all. Is it possible to extend this example, or have an alternate way to achieve this goal, but also have CGI support? The final goal is to have a quick and lazy way of serving a web site from a certain directory. The site has static content (HTML pages, images), but also a CGI script. The CGI script must work properly when accessed via browser. Of course I could setup a virtual host in apache, allow CGI inside it etc. But that's not a zero-config approach.

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  • C++ double division by 0.0 versus DBL_MIN

    - by wonsungi
    When finding the inverse square root of a double, is it better to clamp invalid non-positive inputs at 0.0 or MIN_DBL? (In my example below double b may end up being negative due to floating point rounding errors and because the laws of physics are slightly slightly fudged in the game.) Both division by 0.0 and MIN_DBL produce the same outcome in the game because 1/0.0 and 1/DBL_MIN are effectively infinity. My intuition says MIN_DBL is the better choice, but would there be any case for using 0.0? Like perhaps sqrt(0.0), 1/0.0 and multiplication by 1.#INF000000000000 execute faster because they are special cases. double b = 1 - v.length_squared()/(c*c); #ifdef CLAMP_BY_0 if (b < 0.0) b = 0.0; #endif #ifdef CLAMP_BY_DBL_MIN if (b <= 0.0) b = DBL_MIN; #endif double lorentz_factor = 1/sqrt(b); double division in MSVC: 1/0.0 = 1.#INF000000000000 1/DBL_MIN = 4.4942328371557898e+307

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  • Division inaccurate in Javascript?

    - by Nate
    If I perform the following operation in Javascript: 0.06120*400 The result is 24.48. However, if I do this: 24.48/400 The result is: 0.061200000000000004 JSFiddle: http://jsfiddle.net/zcDH7/ So it appears that Javascript rounds things differently when doing division and multiplication? Using my calculator, the operation 24.48/400 results in the correct answer of 0.0612. How should I deal with Javascript's inaccurate division? I can't simply round the number off, because I will be dealing with numbers of varying precision. Thanks for your advice.

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  • Using multiplication and division with delta time

    - by tesselode
    Using delta time with addition and subtraction is easy. player.x += 100 * dt However, multiplication and division complicate things a bit. For example, let's say I want the player to double his speed every second. player.x = player.x * 2 * dt I can't do this because it'll slow down the player (unless delta time is really high). Division is the same way, except it'll speed things way up. How can I handle multiplication and division with delta time?

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  • How to catch a division by zero?

    - by Cristian Castiblanco
    I have a large mathematical expression that has to be created dinamically. So, for example, once I have parsed "something" the result will be a string like: "$foo+$bar/$baz";. So, for calculating the result of that expression I'm using the eval function... something like this: eval("\$result = $expresion;"); echo "The result is: $result"; The problem here is that sometimes I get errors that says there was a division by zero, and I don't know how to catch that Exception. I have tried things like: eval("try{\$result = $expresion;}catch(Exception \$e){\$result = 0;}"); echo "The result is: $result"; Or: try{ eval("\$result = $expresion;"); } catch(Exception $e){ $result = 0; } echo "The result is: $result"; But it does not work. So, how can I avoid that my application crashes when there is a division by zero?

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  • Polynomial division overloading operator

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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  • VS C++ throwing divide by zero exception after a specific check

    - by Dr. Monkey
    In the following C++ code, it should be impossible for ain integer division by zero to occur: // gradedUnits and totalGrades are both of type int if (gradedUnits == 0) { return 0; } else { return totalGrades/gradedUnits; //call stack points to this line } however Visual Studio is popping up this error: Unhandled exception at 0x001712c0 in DSA_asgn1.exe: 0xC0000094: Integer division by zero. And the stack trace points to the line indicated in the code. It seems like VS might just do this with any integer division, without checking whether a divide by zero is possible. Do I need to catch this exception even though the code should never be able to throw it? If so, what's the best way to go about this? This is for an assignment that specifies VS 2005/2008 with C++. I would prefer not to make things more complicated than I need to, but at the same time I like to do things properly where possible.

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  • Faster integer division when denominator is known?

    - by aaa
    hi I am working on GPU device which has very high division integer latency, several hundred cycles. I am looking to optimize divisions. All divisions by denominator which is in a set { 1,3,6,10 }, however numerator is a runtime positive value, roughly 32000 or less. due to memory constraints, lookup table is not option. Can you think of alternatives? I have thought of computing float point inverses, and using those to multiply numerator. Thanks

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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  • Division, Remainders and only Real Numbers Allowed

    - by Senica Gonzalez
    Trying to figure out this pseudo code. The following is assumed.... I can only use unsigned and signed integers (or long). Division returns a real number with no remainder. MOD returns a real number. Fractions and decimals are not handled. INT I = 41828; INT C = 15; INT D = 0; D = (I / 65535) * C; How would you handle a fraction (or decimal value) in this situation? Is there a way to use negative value to represent the remainder? In this example I/65535 should be 0.638, however, with the limitations, I get 0 with a MOD of 638. How can I then multiply by C to get the correct answer? Hope that makes sense.

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  • What division operator symbol would you pick?

    - by Mackenzie
    I am currently designing and implementing a small programming language as an extra-credit project in a class I'm taking. My problem is that the language has three numeric types: Long, Double, and Fraction. Fractions can be written in the language as proper or improper fractions (e.g. "2 1/3" or "1/2"). This fact leads to problems such as "2/3.5" (Long/Double) and "2/3"(Long/Long) not being handled correctly by the lexer.The best solution that I see is to change the division operator. So far, I think "\" is the best solution since "//" starts comments. Would you pick "\", if you were designing the language? Would you pick something else? If so, what? Note: changing the way fractions are written is not possible. Thanks in advance for your help,

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  • Custom "Very Long Int" Division Issue

    - by befall
    Hey everyone, So, for a very silly project in C++, we are making our own long integer class, called VLI (Very Long Int). The way it works (they backboned it, blame them for stupidity) is this: User inputs up to 50 digits, which are input as string. String is stored in pre-made Sequence class, which stores the string in an array, in reverse order. That means, when "1234" is input, it gets stored as [4|3|2|1]. So, my question is this: How can I go about doing division using only these arrays of chars? If the input answer is over 32 digits, I can't use ints to check for stuff, and they basically saying using long ints here is cheating. Any input is welcome, and I can give more clarification if need be, thanks everyone.

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  • Windows Azure: Server and Cloud Division

    - by kaleidoscope
    On 8th Dec 2009 Microsoft announced the formation of a new organization within the Server & Tools Business that combines the Windows Server & Solutions group and the Windows Azure group, into a single organization called the Server & Cloud Division (SCD). SCD will deliver solutions that help our customers realize even greater benefits from Microsoft’s investments in on-premises and cloud technologies.  And the new division will help strengthen an already solid and extensive partner ecosystem. Together, Windows Server, Windows Azure, SQL Server, SQL Azure, Visual Studio and System Center help customers extend existing investments to include a future that will combine both on-premises and cloud solutions, and SCD is now a key player in that effort. http://blogs.technet.com/windowsserver/archive/2009/12/08/windows-server-and-windows-azure-come-together-in-a-new-stb-organization-the-server-cloud-division.aspx   Tinu, O

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  • integer division properties

    - by aaa
    hi. does the following integer arithmetic property hold? (m/n)/l == m/(n*l) At first I thought I knew answer (does not hold), but now am not sure. Does it hold for all numbers or only for certain conditions, i.e. n > l?

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  • Java - simple division in Java ---> bug/feature?!

    - by msr
    Hello, Im astonished. Im trying this simple calculation in a Java application: System.out.println("b=" + (1 - 7/10)); Obviously Im wainting for "b=0.3" in the output but here's what I get: b=1 What?! Why this happens? If I make: System.out.println("b=" + (1-0.7)); I get the right result which is "b=0.3". What's going wrong here? Thanks!

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  • Python - Number of Significant Digits in results of division

    - by russ
    Newbie here. I have the following code: myADC = 128 maxVoltage = 5.0 maxADC = 255.0 VoltsPerADC = maxVoltage/maxADC myVolts = myADC * VoltsPerADC print "myADC = {0: >3}".format(myADC) print "VoltsPerADC = {0: >7}".format(VoltsPerADC) print VoltsPerADC print "myVolts = {0: >7}".format(myVolts) print myVolts This outputs the following: myADC = 128 VoltsPerADC = 0.0196078 0.0196078431373 myVolts = 2.5098 2.50980392157 I have been searching for an explanation of how the number of significant digits is determined by default, but have had trouble locating an explanation that makes sense to me. This link link text suggests that by default the "print" statement prints numbers to 10 significant figures, but that does not seem to be the case in my results. How are the number of significant digits/precision determined? Can someone shed some light on this for me. Thanks in advance for your time and patience.

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  • Java - simple division in Java ---> bug?!

    - by msr
    Hello, Im astonished. Im trying this simple calculation in a Java application: System.out.println("b=" + (1 - 7/10)); Obviously Im wainting for "b=0.3" in the output but here's what I get: b=1 What?! Why this happens? If I make: System.out.println("b=" + (1-0.7)); I get the right result which is "b=0.3". What's going wrong here? Thanks!

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  • division with wrong result

    - by PeterK
    Hi, I am trying to divide integers but get 0 as result. I just do not understand what i am doing wrong. I am using only int's in this example but get the same result testing with float or double. The code i use is: int wrongAnswers = askedQuestions - playerResult; int percentCorrect = (playerResult / askedQuestions) * 100; int percentWrong = (wrongAnswers / askedQuestions) * 100; NSLog(@"askedQuestions: %i", askedQuestions); NSLog(@"playerResult: %i", playerResult); NSLog(@"wrongAnswers: %i", wrongAnswers); NSLog(@"percentCorrect: %i", percentCorrect); NSLog(@"percentWrong: %i", percentWrong); NSLog(@"calc: %i", (wrongAnswers + playerResult)); NSLog(@"wrong answers %: %i %%", ((wrongAnswers / askedQuestions) * 100)); The result i get is: 2011-01-09 16:45:53.411 XX[8296:207] askedQuestions: 5 2011-01-09 16:45:53.412 XX[8296:207] playerResult: 2 2011-01-09 16:45:53.412 XX[8296:207] wrongAnswers: 3 2011-01-09 16:45:53.413 XX[8296:207] percentCorrect: 0 % 2011-01-09 16:45:53.414 XX[8296:207] percentWrong: 0 % 2011-01-09 16:45:53.414 XX[8296:207] calc: 5 2011-01-09 16:45:53.415 XX[8296:207] wrong answers : 0 % I would very much appreciate help :-)

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