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  • Avoiding new operator in JavaScript -- the better way

    - by greengit
    Warning: This is a long post. Let's keep it simple. I want to avoid having to prefix the new operator every time I call a constructor in JavaScript. This is because I tend to forget it, and my code screws up badly. The simple way around this is this... function Make(x) { if ( !(this instanceof arguments.callee) ) return new arguments.callee(x); // do your stuff... } But, I need this to accept variable no. of arguments, like this... m1 = Make(); m2 = Make(1,2,3); m3 = Make('apple', 'banana'); The first immediate solution seems to be the 'apply' method like this... function Make() { if ( !(this instanceof arguments.callee) ) return new arguments.callee.apply(null, arguments); // do your stuff } This is WRONG however -- the new object is passed to the apply method and NOT to our constructor arguments.callee. Now, I've come up with three solutions. My simple question is: which one seems best. Or, if you have a better method, tell it. First – use eval() to dynamically create JavaScript code that calls the constructor. function Make(/* ... */) { if ( !(this instanceof arguments.callee) ) { // collect all the arguments var arr = []; for ( var i = 0; arguments[i]; i++ ) arr.push( 'arguments[' + i + ']' ); // create code var code = 'new arguments.callee(' + arr.join(',') + ');'; // call it return eval( code ); } // do your stuff with variable arguments... } Second – Every object has __proto__ property which is a 'secret' link to its prototype object. Fortunately this property is writable. function Make(/* ... */) { var obj = {}; // do your stuff on 'obj' just like you'd do on 'this' // use the variable arguments here // now do the __proto__ magic // by 'mutating' obj to make it a different object obj.__proto__ = arguments.callee.prototype; // must return obj return obj; } Third – This is something similar to second solution. function Make(/* ... */) { // we'll set '_construct' outside var obj = new arguments.callee._construct(); // now do your stuff on 'obj' just like you'd do on 'this' // use the variable arguments here // you have to return obj return obj; } // now first set the _construct property to an empty function Make._construct = function() {}; // and then mutate the prototype of _construct Make._construct.prototype = Make.prototype; eval solution seems clumsy and comes with all the problems of "evil eval". __proto__ solution is non-standard and the "Great Browser of mIsERY" doesn't honor it. The third solution seems overly complicated. But with all the above three solutions, we can do something like this, that we can't otherwise... m1 = Make(); m2 = Make(1,2,3); m3 = Make('apple', 'banana'); m1 instanceof Make; // true m2 instanceof Make; // true m3 instanceof Make; // true Make.prototype.fire = function() { // ... }; m1.fire(); m2.fire(); m3.fire(); So effectively the above solutions give us "true" constructors that accept variable no. of arguments and don't require new. What's your take on this. -- UPDATE -- Some have said "just throw an error". My response is: we are doing a heavy app with 10+ constructors and I think it'd be far more wieldy if every constructor could "smartly" handle that mistake without throwing error messages on the console.

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  • PostGres Error When Using Distinct : postgres ERROR: could not identify an ordering operator for ty

    - by CaffeineIV
    ** EDIT ** Nevermind, just needed to take out the parens... I get this error: ERROR: could not identify an ordering operator for type record when trying to use DISTINCT Here's the query: select DISTINCT(g.fielda, g.fieldb, r.type) from fields g LEFT JOIN types r ON g.id = r.id; And the errors: ERROR: could not identify an ordering operator for type record HINT: Use an explicit ordering operator or modify the query. ********** Error ********** ERROR: could not identify an ordering operator for type record SQL state: 42883 Hint: Use an explicit ordering operator or modify the query.

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  • Recognizing when to use the mod operator

    - by Will
    I have a quick question about the mod operator. I know what it does; it calculates the remainder of a division. My question is, how can I identify a situation where I would need to use the mod operator? I know I can use the mod operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the mod operator is useful and I would like to learn to take advantage of it. I just have problems identifying where the mod operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "hey! the remainder of division would work here!" Any tips or strategies? Thanks

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  • Is there an exponent operator in C#?

    - by Charlie
    For example, does an operator exist to handle this? float Result, Number1, Number2; Number1 = 2; Number2 = 2; Result = Number1 (operator) Number2; In the past the ^ operator has served as an exponential operator in other languages, but in C# it is a bit-wise operator. Do I have to write a loop or include another namespace to handle exponential operations? If so, how do I handle exponential operations using non-integers?

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  • Consistency in placing operator functions

    - by wrongusername
    I have a class like this: class A { ...private functions, variables, etc... public: ...some public functions and variables... A operator * (double); A operator / (double); A operator * (A); ...and lots of other operators } However, I want to also be able to do stuff like 2 * A instead of only being allowed to do A * 2, and so I would need functions like these outside of the class: A operator * (double, A); A operator / (double, A); ...etc... Should I put all these operators outside of the class for consistency, or should I keep half inside and half outside?

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  • RAR Password recovery / hash extraction on Mac OS X

    - by Josh K
    I'm running Mac OS X 10.6.2 and have been handed a couple of old files that need to be extracted. Old backups or finances or bills I believe. They are RAR files, and password protected. Is there a way to extract the hash from these files so I can feed it into John The Ripper or Cain and Abel? Edit I have downloaded cRARk, but unfortunately nothing I have (SimplyRAR, RAR Expander, The Unarchiver) will extract it without a password. Can someone verify that I'm crazy and there is no password on the Mac version?

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  • Microsoft ISA 2006 Log Extraction Issues

    - by Johandk
    Hi, We have a client running ISA2006 with logs going to MSDE. The MSDE database gets rotated. We have NO control over the client infrastructure. We need to get the logs out to a TEXT file on the server itself which will be forwarded to a remote server by other software we already have in place. My first thought was doing it via Stored Procs. Another option is vbscript? The actual table holding the logs has no primary key or other identifier so I think the dump should be done once a day for the previous day, but ideally it should run several times a day. Any ideas?

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  • RAR Password recovery / hash extraction on OS X

    - by Josh K
    I'm running 10.6.2 and have been handed a couple of old files that need to be extracted. Old backups or finances or bills I believe. They are RAR files, and password protected. Is there a way to extract the hash from these files so I can feed it into John The Ripper or Cain and Abel? Edit I have downloaded cRARk, but unfortunately nothing I have (SimplyRAR, RAR Expander, The Unarchiver) will extract it without a password. Can someone verify that I'm crazy and there is no password on the Mac version?

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  • Directory tree in a Resource without extraction...

    - by Corelgott
    Hi all, i am looking for a way to store a complete directory including sub directories in an application's resource and not have to extract it to use it. Details: We would like to use GeckoFx (Gecko as C# Component) in one of our applications. GeckoFX needs the XUL-Runner and needs to find it's folder structure We have some other data which I would not prefer to extracted to the customer's pc; At least not onto something persistent like a hdd... Getting the complete directory into the resources is not that kind of a big deal. Compress to one file and done. But not writing it to the disk to use it is something else. I have a strong dislike against temp folders and such things. Would anything like a RAM drive be possible? Some part of the RAM beeing mounted? Does something like this even exist as a lib, or would this only be possible by a device driver? Any thoughts on this? Thanks in advance! Corelgott

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  • Windows command line built-in compression/extraction tool?

    - by Will Marcouiller
    I need to write a batch file to unzip files to their current folder from a given root folder. Folder 0 |----- Folder 1 | |----- File1.zip | |----- File2.zip | |----- File3.zip | |----- Folder 2 | |----- File4.zip | |----- Folder 3 |----- File5.zip |----- FileN.zip So, I wish that my batch file is launched like so: ocd.bat /d="Folder 0" Then, make it iterate from within the batch file through all of the subfolders to unzip the files exactly where the .zip files are located. So here's my question: Does the Windows (from XP at least) have a command line for its embedded zip tool? Otherwise, shall I stick to another third-party util?

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  • CHM e-book cover extraction

    - by Robottinosino
    I would like to extract a "book cover image" from a series of CHM files, which command line tool can do that, please? I have looked at the source of the QuickLook plug-in "quickchm1.1_source" and at the calibre command line conversion options (ebook-convert) but couldn't find anything akin to simple: name_of_tool file_to_convert output_image as in: mytool routermanual.chm routermanualcoverimage.png

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  • Implementing a non-public assignment operator with a public named method?

    - by Casey
    It is supposed to copy an AnimatedSprite. I'm having second thoughts that it has the unfortunate side effect of changing the *this object. How would I implement this feature without the side effect? EDIT: Based on new answers, the question should really be: How do I implement a non-public assignment operator with a public named method without side effects? (Changed title as such). public: AnimatedSprite& AnimatedSprite::Clone(const AnimatedSprite& animatedSprite) { return (*this = animatedSprite); } protected: AnimatedSprite& AnimatedSprite::operator=(const AnimatedSprite& rhs) { if(this == &rhs) return *this; destroy_bitmap(this->_frameImage); this->_frameImage = create_bitmap(rhs._frameImage->w, rhs._frameImage->h); clear_bitmap(this->_frameImage); this->_frameDimensions = rhs._frameDimensions; this->CalcCenterFrame(); this->_frameRate = rhs._frameRate; if(rhs._animation != nullptr) { delete this->_animation; this->_animation = new a2de::AnimationHandler(*rhs._animation); } else { delete this->_animation; this->_animation = nullptr; } return *this; }

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  • What is the rationale to non allow overloading of C++ conversions operator with non-member functio

    - by Vicente Botet Escriba
    C++0x has added explicit conversion operators, but they must always be defined as members of the Source class. The same applies to the assignment operator, it must be defined on the Target class. When the Source and Target classes of the needed conversion are independent of each other, neither the Source can define a conversion operator, neither the Target can define a constructor from a Source. Usually we get it by defining a specific function such as Target ConvertToTarget(Source& v); If C++0x allowed to overload conversion operator by non member functions we could for example define the conversion implicitly or explicitly between unrelated types. template < typename To, typename From operator To(const From& val); For example we could specialize the conversion from chrono::time_point to posix_time::ptime as follows template < class Clock, class Duration operator boost::posix_time::ptime( const boost::chrono::time_point& from) { using namespace boost; typedef chrono::time_point time_point_t; typedef chrono::nanoseconds duration_t; typedef duration_t::rep rep_t; rep_t d = chrono::duration_cast( from.time_since_epoch()).count(); rep_t sec = d/1000000000; rep_t nsec = d%1000000000; return posix_time::from_time_t(0)+ posix_time::seconds(static_cast(sec))+ posix_time::nanoseconds(nsec); } And use the conversion as any other conversion. So the question is: What is the rationale to non allow overloading of C++ conversions operator with non-member functions?

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  • What's the false operator in C# good for?

    - by Jakub Šturc
    There are two weird operators in C#: the true operator the false operator If I understand this right these operators can be used in types which I want to use instead of a boolean expression and where I don't want to provide an implicit conversion to bool. Let's say I have a following class: public class MyType { public readonly int Value; public MyType(int value) { Value = value; } public static bool operator true (MyType mt) { return mt.Value > 0; } public static bool operator false (MyType mt) { return mt.Value < 0; } } So I can write the following code: MyType mTrue = new MyType(100); MyType mFalse = new MyType(-100); MyType mDontKnow = new MyType(0); if (mTrue) { // Do something. } while (mFalse) { // Do something else. } do { // Another code comes here. } while (mDontKnow) However for all the examples above only the true operator is executed. So what's the false operator in C# good for? Note: More examples can be found here, here and here.

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  • Why do not C++11's move constructor/assignment operator act as expected

    - by xmllmx
    #include <iostream> using namespace std; struct A { A() { cout << "A()" << endl; } ~A() { cout << "~A()" << endl; } A(A&&) { cout << "A(A&&)" << endl; } A& operator =(A&&) { cout << "A& operator =(A&&)" << endl; return *this; } }; struct B { // According to the C++11, the move ctor/assignment operator // should be implicitly declared and defined. The move ctor // /assignment operator should implicitly call class A's move // ctor/assignment operator to move member a. A a; }; B f() { B b; // The compiler knows b is a temporary object, so implicitly // defined move ctor/assignment operator of class B should be // called here. Which will cause A's move ctor is called. return b; } int main() { f(); return 0; } My expected output should be: A() A(A&&) ~A() ~A() However, the actual output is: (The C++ compiler is: Visual Studio 2012) A() ~A() ~A() Is this a bug of VC++? or just my misunderstanding?

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  • C++ operator lookup rules / Koenig lookup

    - by John Bartholomew
    While writing a test suite, I needed to provide an implementation of operator<<(std::ostream&... for Boost unit test to use. This worked: namespace theseus { namespace core { std::ostream& operator<<(std::ostream& ss, const PixelRGB& p) { return (ss << "PixelRGB(" << (int)p.r << "," << (int)p.g << "," << (int)p.b << ")"); } }} This didn't: std::ostream& operator<<(std::ostream& ss, const theseus::core::PixelRGB& p) { return (ss << "PixelRGB(" << (int)p.r << "," << (int)p.g << "," << (int)p.b << ")"); } Apparently, the second wasn't included in the candidate matches when g++ tried to resolve the use of the operator. Why (what rule causes this)? The code calling operator<< is deep within the Boost unit test framework, but here's the test code: BOOST_AUTO_TEST_SUITE(core_image) BOOST_AUTO_TEST_CASE(test_output) { using namespace theseus::core; BOOST_TEST_MESSAGE(PixelRGB(5,5,5)); // only compiles with operator<< definition inside theseus::core std::cout << PixelRGB(5,5,5) << "\n"; // works with either definition BOOST_CHECK(true); // prevent no-assertion error } BOOST_AUTO_TEST_SUITE_END() For reference, I'm using g++ 4.4 (though for the moment I'm assuming this behaviour is standards-conformant).

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  • On C++ global operator new: why it can be replaced

    - by Jimmy
    I wrote a small program in VS2005 to test whether C++ global operator new can be overloaded. It can. #include "stdafx.h" #include "iostream" #include "iomanip" #include "string" #include "new" using namespace std; class C { public: C() { cout<<"CTOR"<<endl; } }; void * operator new(size_t size) { cout<<"my overload of global plain old new"<<endl; // try to allocate size bytes void *p = malloc(size); return (p); } int main() { C* pc1 = new C; cin.get(); return 0; } In the above, my definition of operator new is called. If I remove that function from the code, then operator new in C:\Program Files (x86)\Microsoft Visual Studio 8\VC\crt\src\new.cpp gets called. All is good. However, in my opinion, my implementations of operator new does NOT overload the new in new.cpp, it CONFLICTS with it and violates the one-definition rule. Why doesn't the compiler complain about it? Or does the standard say since operator new is so special, one-definition rule does not apply here? Thanks.

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  • Operator+ for a subtype of a template class.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<int>::subtype operator+(const c<int>::subtype& left, // const c<int>::subtype& right) // { return c<int>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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  • Operator+ for a subtype of a template classe.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<a>::subtype operator+(const c<a>::subtype& left, // const c<a>::subtype& right) // { return c<a>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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