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  • operator overloading

    - by cpp_Beginner
    Hi, Could anybody tell me the difference between operator overloading using the friend keyword and as a member function inside a class? also what is the difference incase of any unary operator overloading i.e., as a friend and as a member function

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  • Negate the null-coalescing operator

    - by jhunter
    I have a bunch of strings I need to use .Trim() on, but they can be null. It would be much more concise if I could do something like: string endString = startString !?? startString.Trim(); Basically return the part on the right if the part on the left is NOT null, otherwise just return the null value. I just ended up using the ternary operator, but is there anyway to use the null-coalescing operator for this purpose?

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  • behaviour of the implicit copy constructor / assignment operator

    - by Tobias Langner
    Hello, I have a question regarding the C++ Standard. Suppose you have a base class with user defined copy constructor and assignment operator. The derived class uses the implicit one generated by the compiler. Does copying / assignment of the derived class call the user defined copy constructor / assignment operator? Or do you need to implement user defined versions that call the base class? Thank you for your help.

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  • C++ segmentation error when first parameter is null in comparison operator overload

    - by user1774515
    I am writing a class called Word, that handles a c string and overloads the <, , <=, = operators. word.h: friend bool operator<(const Word &a, const Word &b); word.cc: bool operator<(const Word &a, const Word &b) { if(a == NULL && b == NULL) return false; if(a == NULL) return true; if(b == NULL) return false; return a.wd < b.wd; //wd is a valid c string } main: char* temp = NULL; //EDIT: i was mistaken, temp is a char pointer Word a("blah"); //a.wd = [b,l,a,h] cout << (temp<a); i get a segmentation error before the first line of the operator< method after the last line in the main. I can correct the problem by writing cout << (a>temp); where the operator> is similarly defined and i get no errors. but my assignment requires (temp < a) to work so this is where i ask for help. EDIT: i made a mistake the first time and i said temp was of type Word, but it is actually of type char*. so i assume that the compiler converts temp to a Word using one of my constructors. i dont know which one it would use and why this would work since the first parameter is not Word. here is the constructor i think is being used to make the Word using temp: Word::Word(char* c, char* delimeters=NULL) { char *temporary = "\0"; if(c == NULL) c = temporary; check(stoppers!=NULL, "(Word(char*,char*))NULL pointer"); //exits the program if the expression is false if(strlen(c) == 0) size = DEFAULT_SIZE; //10 else size = strlen(c) + 1 + DEFAULT_SIZE; wd = new char[size]; check(wd!=NULL, "Word(char*,char*))heap overflow"); delimiters = new char[strlen(stoppers) + 1]; //EDIT: changed to [] check(delimiters!=NULL,"Word(char*,char*))heap overflow"); strcpy(wd,c); strcpy(delimiters,stoppers); count = strlen(wd); } wd is of type char* thanks for looking at this big question and trying to help. let me know if you need more code to look at

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  • What is operator<< <> in C++?

    - by Austin Hyde
    I have seen this in a few places, and to confirm I wasn't crazy, I looked for other examples. Apparently this can come in other flavors as well, eg operator+ <>. However, nothing I have seen anywhere mentions what it is, so I thought I'd ask. It's not the easiest thing to google operator<< <>( :-)

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  • Why can operator-> be overloaded manually?

    - by FredOverflow
    Wouldn't it make sense if p->m was just syntactic sugar for (*p).m? Essentially, every operator-> that I have ever written could have been implemented as follows: Foo::Foo* operator->() { return &**this; } Is there any case where I would want p->m to mean something else than (*p).m?

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  • How to push_back without operator=() for const members?

    - by WilliamKF
    How to push_back() to a C++ std::vector without using operator=() for which the default definition violates having const members? struct Item { Item(int value) : _value(value) { } const int _value; } vector<Item> items; items.push_back(Item(3)); I'd like to keep the _value const since it should not change after the object is constructed, so the question is how do I initialize my vector with elements without invoking operator=()?

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  • Two new profile in new visual studio 2010.

    - by Jalpesh P. Vadgama
    Visual studio 2010 is a great tool and i have become fan of visual studio 2010. I have found two new code profile in visual studio 2010. Web Development Profile Web Development Code Optimized Profile. Web Development profile will hide the top bar which contains the client object and and event dropdowns. So it will have more spaces. Another one web development code optimized which will hide all the things except main windows. It will hide Toolbox,CSS properties and all other things so you will have more spaces to play with your html. So as a web developer you can use this two great new profile as per your convenience when you only want to play with your html then use webdevelopement code  optimized profile and another interesting thing is that you don’t have to reset your settings you can also just do with Tools->Settings menu like below. This will swap different profile like below. Hope this will help you.. Technorati Tags: Visual Studio 2010,ASP.NET 4.0

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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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  • 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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  • New EBS 12.0 AutoConfig Rollup 7 Now Available

    - by Steven Chan
    AutoConfig manages the configuration of E-Business Suite environments.  The seventh and latest rollup patch for the AutoConfig engine and tools for Oracle E-Business Suite Release 12.0 is now available for download.  The official (and admittedly-cryptic) name for this EBS 12.0 patch is: R12.TXK.A.DELTA.7 (Patch 9386653)

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  • Discoverer 11g 11.1.1.2 Certified with EBS 12 on Five New Platforms

    - by Steven Chan
    Oracle Fusion Middleware 11g Release 1 includes Oracle Discoverer.  Discoverer is an ad-hoc query, reporting, analysis, and Web-publishing tool that allows end-users to work directly with Oracle E-Business Suite OLTP data.We certified Discoverer 11gR1 11.1.1.2 with the E-Business Suite Release 11i and 12 on Linux earlier this year.  Our Applications Platforms Group has just released five additional platform certifications for Discoverer 11.1.1.2 for Oracle E-Business Suite Release 12 (12.0.x and 12.1.x).Certified EBS 12 PlatformsLinux x86-64 (Oracle Enterprise Linux 4, 5) Linux x86-64 (RHEL 4, 5) Linux x86-64 (SLES 10) Oracle Solaris on SPARC (64-bit) (Solaris 9, 10) HP-UX Itanium (11.23, 11.31) HP-UX PA-RISC (64-bit) (11.23, 11.31) IBM AIX on Power Systems (64-bit) (5.3, 6.1) Microsoft Windows Server (32-bit) (2003, 2008)

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  • Dealing with a fundamental design flaw when you're new to the project

    - by Matt Phillips
    I've just started working on an open source project with around 30 developers in it. I'm working on fixing some of the bugs as a way to get into the "loop" and become a regular committer to the project. The problem is I think I've uncovered a fundamental design flaw that's causing one of the bugs I'm working on. But I feel like if I blast this on the mailing list I'm going to come off as arrogant, and some of the discussions I've had about the issue are butting heads with some of the people. How should I go about this?

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  • New Wine in New Bottles

    - by Tony Davis
    How many people, when their car shows signs of wear and tear, would consider upgrading the engine and keeping the shell? Even if you're cash-strapped, you'll soon work out the subtlety of the economics, the cost of sudden breakdowns, the precious time lost coping with the hassle, and the low 'book value'. You'll generally buy a new car. The same philosophy should apply to database systems. Mainstream support for SQL Server 2005 ends on April 12; many DBAS, if they haven't done so already, will be considering the migration to SQL Server 2008 R2. Hopefully, that upgrade plan will include a fresh install of the operating system on brand new hardware. SQL Server 2008 R2 and Windows Server 2008 R2 are designed to work together. The improved architecture, processing power, and hyper-threading capabilities of modern processors will dramatically improve the performance of many SQL Server workloads, and allow consolidation opportunities. Of course, there will be many DBAs smiling ruefully at the suggestion of such indulgence. This is nothing like the real world, this halcyon place where hardware and software budgets are limitless, development and testing resources are plentiful, and third party vendors immediately certify their applications for the latest-and-greatest platform! As with cars, or any other technology, the justification for a complete upgrade is complex. With Servers, the extra cost at time of upgrade will generally pay you back in terms of the increased performance of your business applications, reduced maintenance costs, training costs and downtime. Also, if you plan and design carefully, it's possible to offset hardware costs with reduced SQL Server licence costs. In his forthcoming SQL Server Hardware book, Glenn Berry describes a recent case where he was able to replace 4 single-socket database servers with one two-socket server, saving about $90K in hardware costs and $350K in SQL Server license costs. Of course, there are exceptions. If you do have a stable, reliable, secure SQL Server 6.5 system that still admirably meets the needs of a specific business requirement, and has no security vulnerabilities, then by all means leave it alone. Why upgrade just for the sake of it? However, as soon as a system shows sign of being unfit for purpose, or is moving out of mainstream support, the ruthless DBA will make the strongest possible case for a belts-and-braces upgrade. We'd love to hear what you think. What does your typical upgrade path look like? What are the major obstacles? Cheers, Tony.

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  • [New England] SQL Saturday 71 - April 2 - Boston Area

    - by Adam Machanic
    April in the Boston area means many things. The Boston Marathon, the beginning of baseball season, and -- hopefully -- a bit of a respite from the ridiculously cold and snowy winter we've been having. This April will mean one more thing: A full-day, free SQL Server event featuring 30 top-notch sessions . SQL Saturday 71 will be the third full-day event in the area in as many years, and is shaping up to be the best yet. For the past several months I've been working and planning in conjunction with...(read more)

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  • Looking for new language and new technology [closed]

    - by Basim
    back when Microsoft relased .Net in 2002 or whatever, when I look at that time I say to myself what I if I picked one of Microsoft language in that time and still work on it, of course I will be professional by now. I am looking for a new language that is going up and will be big thing in the next 5-10 years, so in that time i can see the big picture and I know that I'm one of the few people who started from the beginning with X programming language or technology. My interest is web development.

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  • advice for a new software engineer/developer right out of college

    - by ranzy
    I just graduated recently from a 4-year university with a degree in Computer Science and thankfully got a job in Software Engineering. I'm working with C++ with a .NET framework if that's correct to say because that also confuses me. What I'm asking for is what tutorials/books are out there to learn C++ for Windows Programming I guess? I know how to program somewhat and I understand the concepts but when I look at the code it doesn't make sense to me. I know I just started so it's kind of expected but it's certainly quite different from college. Thanks!

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  • How to install new system (with new applications) without removing /home

    - by Innuendo
    I'd like to update to 11.04 from 10.10 (but I don't like upgrading system, I prefer full reinstalling) I'd like to install whole new system, but I want to keep my /home folders (Music, Movies, Documents and so on). If I keep /home while installing - this will keep all my program settings too (but I have lots of trash there too, and 'd like to clean it too). Can I delete all .folders in /home and then reinstall system (keeping old /home) ?

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  • 3x3 Sobel operator and gradient features

    - by pithyless
    Reading a paper, I'm having difficulty understanding the algorithm described: Given a black and white digital image of a handwriting sample, cut out a single character to analyze. Since this can be any size, the algorithm needs to take this into account (if it will be easier, we can assume the size is 2^n x 2^m). Now, the description states given this image we will convert it to a 512-bit feature (a 512-bit hash) as follows: (192 bits) computes the gradient of the image by convolving it with a 3x3 Sobel operator. The direction of the gradient at every edge is quantized to 12 directions. (192 bits) The structural feature generator takes the gradient map and looks in a neighborhood for certain combinations of gradient values. (used to compute 8 distinct features that represent lines and corners in the image) (128 bits) Concavity generator uses an 8-point star operator to find coarse concavities in 4 directions, holes, and lagrge-scale strokes. The image feature maps are normalized with a 4x4 grid. I'm for now struggling with how to take an arbitrary image, split into 16 sections, and using a 3x3 Sobel operator to come up with 12 bits for each section. (But if you have some insight into the other parts, feel free to comment :)

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  • operator<< overload,

    - by mr.low
    //using namespace std; using std::ifstream; using std::ofstream; using std::cout; class Dog { friend ostream& operator<< (ostream&, const Dog&); public: char* name; char* breed; char* gender; Dog(); ~Dog(); }; im trying to overload the << operator. I'm also trying to practice good coding. But my code wont compile unless i uncomment the using namespace std. i keep getting this error and i dont know. im using g++ compiler. Dog.h:20: error: ISO C++ forbids declaration of ‘ostream’ with no type Dog.h:20: error: ‘ostream’ is neither function nor member function; cannot be declared friend. if i add line using std::cout; then i get this error. Dog.h:21: error: ISO C++ forbids declaration of ‘ostream’ with no type. Can somebody tell me the correct way to overload the << operator with out using namespace std;

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  • How to overload operator<< for qDebug

    - by iyo
    Hi, I'm trying to create more useful debug messages for my class where store data. My code is looking something like this #include <QAbstractTableModel> #include <QDebug> /** * Model for storing data. */ class DataModel : public QAbstractTableModel { // for debugging purposes friend QDebug & operator<< (const QDebug &d, DataModel model); //other stuff }; /** * Overloading operator for debugging purposes */ QDebug & operator<< (QDebug &d, DataModel model) { d << "Hello world!"; return d; } I expect qDebug() << model will print "Hello world!". However, there is alway something like "QAbstractTableModel(0x1c7e520)" on the output. Do you have any idea what's wrong?

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