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  • I have to generate PL/SQL using Java. Most of the procedures are common. Only a few keeps changing.

    - by blog
    I have to generate PL-SQL code, with some common code(invariable) and a variable code. I don't want to use any external tools. Some ways that I can think: Can I go and maintain the common code in a template and with markers, where my java code will generate code in the markers and generate a new file. Maintain the common code in static constant String and then generate the whole code in StringBuffer and at last write to file. But, I am not at all satisfied with both the ideas. Can you please suggest any better ways of doing this or the use of any design patterns or anything? Thanks in Advance.

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  • Using HTML::Template within a value attribute

    - by Zerobu
    Hello, my question is how would I use an HTML::Template tag inside a value of form to change that form. For example <table border="0" cellpadding="8" cellspacing="1"> <tr> <td align="right">File:</td> <td> <input type="file" name="upload" value= style="width:400px"> </td> </tr> <tr> <td align="right">File Name:</td> <td> <input type="text" name="filename" style="width:400px" value="" > </td> </tr> <tr> <td align="right">Title:</td> <td> <input type="text" name="title" style="width:400px" value="" /> </td> </tr> <tr> <td align="right">Date:</td> <td> <input type="text" name="date" style="width:400px" value="" /> </td> </tr> <tr> <td colspan="2" align="right"> <input type="button" value="Cancel"> <input type="submit" name="action" value="Upload" /> </td> </tr> </table> I want the value to have a variable in it.

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  • Installing a custom project template with Visual Studio Installer project

    - by ulu
    Hi! I've created a custom project template, and now I need to deploy it together with my product (i.e., it should be installed by the same msi I use for the main installation). I'm using a Visual Studio Installer project. One option is to use a custom action and manually copy a template file included in the installation. Another is to create a vsi file and use a custom action to install it after the main installation (how do I have it installed silently?) . Which one is better? Thanks a lot ulu

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  • Why do I get the error "X is not a member of Y" even though X is a friend of Y?

    - by user1232138
    I am trying to write a binary tree. Why does the following code report error C2039, "'<<' : is not a member of 'btree<T'" even though the << operator has been declared as a friend function in the btree class? #include<iostream> using namespace std; template<class T> class btree { public: friend ostream& operator<<(ostream &,T); }; template<class T> ostream& btree<T>::operator<<(ostream &o,T s) { o<<s.i<<'\t'<<s.n; return o; }

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  • Possible for C++ template to check for a function's existence?

    - by andy
    Is it possible to write a C++ template that changes behavior depending on if a certain member function is defined on a class? Here's a simple example of what I would want to write: template<class T> std::string optionalToString(T* obj) { if (FUNCTION_EXISTS(T->toString)) return obj->toString(); else return "toString not defined"; } So if class T has "toString" defined then it uses it, otherwise it doesn't. The magical part that I don't know how to do is the "FUNCTION_EXISTS" part.

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  • Should this work?

    - by Noah Roberts
    I am trying to specialize a metafunction upon a type that has a function pointer as one of its parameters. The code compiles just fine but it will simply not match the type. #include <iostream> #include <boost/mpl/bool.hpp> #include <boost/mpl/identity.hpp> template < typename CONT, typename NAME, typename TYPE, TYPE (CONT::*getter)() const, void (CONT::*setter)(TYPE const&) > struct metafield_fun {}; struct test_field {}; struct test { int testing() const { return 5; } void testing(int const&) {} }; template < typename T > struct field_writable : boost::mpl::identity<T> {}; template < typename CONT, typename NAME, typename TYPE, TYPE (CONT::*getter)() const > struct field_writable< metafield_fun<CONT,NAME,TYPE,getter,0> > : boost::mpl::false_ {}; typedef metafield_fun<test, test_field, int, &test::testing, 0> unwritable; int main() { std::cout << typeid(field_writable<unwritable>::type).name() << std::endl; std::cin.get(); } Output is always the type passed in, never bool_.

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  • User Defined Class as a Template Parameter

    - by isurulucky
    Hi, I' m implementing a custom STL map. I need to make sure that any data type (basic or user defined) key will work with it. I declared the Map class as a template which has two parameters for the key and the value. My question is if I need to use a string as the key type, how can I overload the < and operators for string type keys only?? In template specialization we have to specialize the whole class with the type we need as I understand it. Is there any way I can do this in a better way?? What if I add a separate Key class and use it as the template type for Key? Thank You!!

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  • Django Getting RequestContext in custom tag

    - by greggory.hz
    I'm trying to create a custom tag. Inside this custom tag, I want to be able to have some logic that checks if the user is logged in, and then have the tag rendered accordingly. This is what I have: class UserActionNode(template.Node): def __init__(self): pass def render(self, context): if context.user.is_authenticated(): return render_to_string('layout_elements/sign_in_register.html'); else: return render_to_string('layout_elements/logout_settings.html'); def user_actions(parser, test): return UserActionNode() register.tag('user_actions', user_actions) When I run this, I get this error: Caught AttributeError while rendering: 'Context' object has no attribute 'user' The view that renders this looks like this: return render_to_response('start/home.html', {}, context_instance=RequestContext(request)) Why doesn't the tag get a RequestContext object instead of the Context object? How can I get the tag to receive the RequestContext instead of the Context? EDIT: Whether or not it's possible to get a RequestContext inside a custom tag, I'd still be interested to know the "correct" or best way to determine a user's authentication state from within the custom tag. If that's not possible, then perhaps that kind of logic belongs elsewhere? Where?

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  • static member specialization of templated child class and templated base class

    - by b3nj1
    I'm trying to have a templated class (here C) that inherits from another templated class (here A) and perform static member specialization (of int var here), but I cant get the right syntax to do so (if it's possible #include <iostream> template<typename derived> class A { public: static int var; }; //This one works fine class B :public A<B> { public: B() { std::cout << var << std::endl; } }; template<> int A<B>::var = 9; //This one doesn't works template<typename type> class C :public A<C<type> > { public: C() { std::cout << var << std::endl; } }; //template<> template<typename type> int A<C<type> >::a = 10; int main() { B b; C<int> c; return 0; } I put an example that works with a non templated class (here B) and i can get the static member specialization of var, but for C that just doesn't work. Here is what gcc tells me : test.cpp: In constructor ‘C<type>::C()’: test.cpp:29:26: error: ‘var’ was not declared in this scope test.cpp: At global scope: test.cpp:34:18: error: template definition of non-template ‘int A<C<type> >::a’ I'm using gcc version 4.6.3, thanks for any help

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  • C++ template member specialization - is this a compiler limitation?

    - by LoudNPossiblyRight
    Is it possible to do this kind of specialization? If so, how? The specialization in question is marked //THIS SPECIALIZATION WILL NOT COMPILE I have used VS2008, VS2010, gcc 4.4.3 and neither can compile this. #include<iostream> #include<string> using namespace std; template <typename ALPHA> class klass{ public: template <typename BETA> void func(BETA B); }; template <typename ALPHA> template <typename BETA> void klass<ALPHA>::func(BETA B){ cout << "I AM A BETA FUNC: " << B <<endl; } //THIS SPECIALIZATION WILL NOT COMPILE template <typename ALPHA> template <> void klass<ALPHA>::func(string B){ cout << "I AM A SPECIAL BETA FUNC: " << B <<endl; } int main(){ klass<string> k; k.func(1); k.func("hello"); return 0; }

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  • Derived template override return type of member function C++

    - by Ruud v A
    I am writing matrix classes. Take a look at this definition: template <typename T, unsigned int dimension_x, unsigned int dimension_y> class generic_matrix { ... generic_matrix<T, dimension_x - 1, dimension_y - 1> minor(unsigned int x, unsigned int y) const { ... } ... } template <typename T, unsigned int dimension> class generic_square_matrix : public generic_matrix<T, dimension, dimension> { ... generic_square_matrix(const generic_matrix<T, dimension, dimension>& other) { ... } ... void foo(); } The generic_square_matrix class provides additional functions like matrix multiplication. Doing this is no problem: generic_square_matrix<T, 4> m = generic_matrix<T, 4, 4>(); It is possible to assign any square matrix to M, even though the type is not generic_square_matrix, due to the constructor. This is possible because the data does not change across children, only the supported functions. This is also possible: generic_square_matrix<T, 4> m = generic_square_matrix<T, 5>().minor(1,1); Same conversion applies here. But now comes the problem: generic_square_matrix<T, 4>().minor(1,1).foo(); //problem, foo is not in generic_matrix<T, 3, 3> To solve this I would like generic_square_matrix::minor to return a generic_square_matrix instead of a generic_matrix. The only possible way to do this, I think is to use template specialisation. But since a specialisation is basically treated like a separate class, I have to redefine all functions. I cannot call the function of the non-specialised class as you would do with a derived class, so I have to copy the entire function. This is not a very nice generic-programming solution, and a lot of work. C++ almost has a solution for my problem: a virtual function of a derived class, can return a pointer or reference to a different class than the base class returns, if this class is derived from the class that the base class returns. generic_square_matrix is derived from generic_matrix, but the function does not return a pointer nor reference, so this doesn't apply here. Is there a solution to this problem (possibly involving an entirely other structure; my only requirements are that the dimensions are a template parameter and that square matrices can have additional functionality). Thanks in advance, Ruud

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  • How to save link with tag e parameters in TextField

    - by xRobot
    I have this simple Post model: class Post(models.Model): title = models.CharField(_('title'), max_length=60, blank=True, null=True) body = models.TextField(_('body')) blog = models.ForeignKey(Blog, related_name="posts") user = models.ForeignKey(User) I want that when I insert in the form the links, the these links are saved in the body from this form: http://www.example.com or www.example.com to this form ( with tag and rel="nofollow" parameter ): <a href="http://www.example.com" rel="nofollow">www.example.com</a> How can I do this ? Thanks ^_^

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  • Constant template parameter class manages to link externally

    - by the_drow
    I have a class foo with an enum template parameter and for some reason it links to two versions of the ctor in the cpp file. enum Enum { bar, baz }; template <Enum version = bar> class foo { public: foo(); }; // CPP File #include "foo.hpp" foo<bar>::foo() { cout << "bar"; } foo<baz>::foo() { cout << "baz"; } I'm using msvc 2008, is this the standard behavior? Are only type template parameters cannot be linked to cpp files?

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  • Does template class/function specialization improves compilation/linker speed?

    - by Stormenet
    Suppose the following template class is heavily used in a project with mostly int as typename and linker speed is noticeably slower since the introduction of this class. template <typename T> class MyClass { void Print() { std::cout << m_tValue << std::endl;; } T m_tValue; } Will defining a class specialization benefit compilation speed? eg. void MyClass<int>::Print() { std::cout << m_tValue << std::endl; }

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  • What does "static" mean in the context of declaring global template functions?

    - by smf68
    I know what static means in the context of declaring global non-template functions (see e.g. What is a "static" function?), which is useful if you write a helper function in a header that is included from several different locations and want to avoid "duplicate definition" errors. So my question is: What does static mean in the context of declaring global template functions? Please note that I'm specifically asking about global, non-member template functions that do not belong to a class. In other words, what is the difference between the following two: template <typename T> void foo(T t) { /* implementation of foo here */ } template <typename T> static void bar(T t) { /* implementation of bar here */ }

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  • Check if a type is an instantiation of a template

    - by Pedro Lacerda
    I have structs like struct RGBA (T) {/* ... */} struct BMPFile (DataT) if (is(DataT == RGBA)) {/* ... */} But is(DataT == RGBA) cannot work because DataT is a type and RGBA is a template. Instead I need check if a type is an instantiation of a template in order to declare file like BMPFile!(RGBA!ushort) file; In a comment @FeepingCreature showed struct RGBA(T) { alias void isRGBAStruct; } struct BMPFile (DataT) if (is(DataT.isRGBAStruct)) {} Although to be working I have no tips on alias void isRGBAStruct.

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  • Specify a base classes template parameters while instantiating a derived class?

    - by DaClown
    Hi, I have no idea if the title makes any sense but I can't find the right words to descibe my "problem" in one line. Anyway, here is my problem. There is an interface for a search: template <typename InputType, typename ResultType> class Search { public: virtual void search (InputType) = 0; virtual void getResult(ResultType&) = 0; }; and several derived classes like: template <typename InputType, typename ResultType> class XMLSearch : public Search<InputType, ResultType> { public: void search (InputType) { ... }; void getResult(ResultType&) { ... }; }; The derived classes shall be used in the source code later on. I would like to hold a simple pointer to a Search without specifying the template parameters, then assign a new XMLSearch and thereby define the template parameters of Search and XMLSearch Search *s = new XMLSearch<int, int>(); I found a way that works syntactically like what I'm trying to do, but it seems a bit odd to really use it: template <typename T> class Derived; class Base { public: template <typename T> bool GetValue(T &value) { Derived<T> *castedThis=dynamic_cast<Derived<T>* >(this); if(castedThis) return castedThis->GetValue(value); return false; } virtual void Dummy() {} }; template <typename T> class Derived : public Base { public: Derived<T>() { mValue=17; } bool GetValue(T &value) { value=mValue; return true; } T mValue; }; int main(int argc, char* argv[]) { Base *v=new Derived<int>; int i=0; if(!v->GetValue(i)) std::cout<<"Wrong type int."<<std::endl; float f=0.0; if(!v->GetValue(f)) std::cout<<"Wrong type float."<<std::endl; std::cout<<i<<std::endl<<f; char c; std::cin>>c; return 0; } Is there a better way to accomplish this?

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  • Printing factorial at compile time in C++

    - by user519882
    template<unsigned int n> struct Factorial { enum { value = n * Factorial<n-1>::value}; }; template<> struct Factorial<0> { enum {value = 1}; }; int main() { std::cout << Factorial<5>::value; std::cout << Factorial<10>::value; } above program computes factorial value during compile time. I want to print factorial value at compile time rather than at runtime using cout. How can we achive printing the factorial value at compile time? I am using VS2009. Thanks!

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  • Template neglects const (why?)

    - by Gabriel
    Does somebody know, why this compiles?? template< typename TBufferTypeFront, typename TBufferTypeBack = TBufferTypeFront> class FrontBackBuffer{ public: FrontBackBuffer( const TBufferTypeFront front, const TBufferTypeBack back): ////const reference assigned to reference??? m_Front(front), m_Back(back) { }; ~FrontBackBuffer() {}; TBufferTypeFront m_Front; ///< The front buffer TBufferTypeBack m_Back; ///< The back buffer }; int main(){ int b; int a; FrontBackBuffer<int&,int&> buffer(a,b); // buffer.m_Back = 33; buffer.m_Front = 55; } I compile with GCC 4.4. Why does it even let me compile this? Shouldn't there be an error that I cannot assign a const reference to a non-const reference?

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  • reverse many to many fields in Django + count them

    - by cleliodpaula
    I'm trying to figure out how to solve this class Item(models.Model): type = models.ForeignKey(Type) name = models.CharField(max_lenght = 10) ... class List(models.Model): items = models.ManyToManyField(Item) ... I want to count how many an Item appears in another Lists, and show on template. view def items_by_list(request, id_): list = List.objects.get(id = id_) qr = list.items.all() #NOT TESTED num = [] i = 0 for item in qr: num[i] = List.objects.filter(items__id = item__id ).count() #FINISH NOT TESTED c = {} c.update(csrf(request)) c = {'request':request, 'list' : qr, 'num' : num} return render_to_response('items_by_list.html', c, context_instance=RequestContext(request)) template {% for dia in list %} <div class="span4" > <div> <h6 style="color: #9937d8">{{item.type.description}}</h6> <small style="color: #b2e300">{{ item.name }}</small> <small style="color: #b2e300">{{COUNT HOW MANY TIMES THE ITEM APPEAR ON OTHER LISTS}}</small> </div> {% endfor %} This seems to be easy, but I could not implement yet. If anyone has some glue to me, please help me. Thanks in advance.

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  • C++ - checking if a class has a certain method at compile time

    - by jetwolf
    Here's a question for the C++ gurus out there. Is there a way to check at compile time where a type has a certain method, and do one thing if it does, and another thing if it doesn't? Basically, I have a template function template <typename T> void function(T t); and I it to behave a certain way if T has a method g(), and another way if it doesn't. Perhaps there is something that can be used together with boost's enable_if? Something like this: template <typename T> enable_if<has_method<T, g, void ()>, void>::type function(T t) { // Superior implementation calling t.g() } template <typename T> disable_if<has_method<T, g, void ()>, void>::type function(T t) { // Inferior implementation in the case where T doesn't have a method g() } "has_method" would be something that preferably checks both that T has a method named 'g', and that the method has the correct signature (in this case, void ()). Any ideas?

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  • How can I have multiple navigation paths with Django, like a simplifies wizard path and a full path?

    - by Zeta
    Lets say I have an application with a structure such as: System set date set name set something Other set death ray target calibrate and I want to have "back" and "next" buttons on a page. The catch is, if you're going in via the "wizard", I want the nav path to be something like "set name" - "set death ray target" - "set name". If you go via the Advanced options menu, I want to just iterate options... "set date" - "set name" - "set something" - "set death ray target" - calibrate. So far, I'm thinking I have to use different URIs, but that's that. Any ideia how this could be done? Thanks.

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