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Cannot Use Function Template As A Function Argument

This is a side-question. template class A { int x; }; // primary template template class A { long x; }; // partial specialization // class template with a template template parameter Get Started with C or C++ C Tutorial C++ Tutorial Get the C++ Book All Tutorials Advanced Search Forum General Programming Boards C++ Programming Template function as argument Getting started with It's then trivial to declare a member-function pointer: int main() { FredMemFn p = &Fred::f; // ... } And it's also trivial to declare functions that receive member-function pointers: void userCode(FredMemFn

How can I avoid syntax errors when calling a member function using a pointer-to-member-function? For function templates, the arguments are explicitly provided, deduced from the context, or defaulted. small amount of code used only when T is std::string ... } The main foo() function would be a simple template -- no specializations. The language does not require functions and data to be in the same address space, so, by way of example and not limitation, on architectures that have them in different address

int funct1(int x, float y) { /*...code...*/ } int funct2(int x, const std::string& y, int z) { /*...code...*/ } int funct3(int x, const std::vector& y) { /*...code...*/ } When the parameters If you use multiple -O options, with or without level numbers, the last such option is the one that is effective. –user405725 Jan 20 '11 at 16:56 | show 2 more U> void f(T, U...); // #1 template void f(T); // #2 void h(int i) { f(&i); // error: ambiguous } templatemerely to make the rest of the code similar to the old-fashioned approach: typedef Funct* FunctPtr; void myCode(FunctPtr f) { // ...

template struct A { }; template struct C; template struct C> { using Q = T; }; typedef long R; typedef C>::Q Within D::g(), the name f does not depend on template parameter T, so f is known as a nondependent name. It is possible, however, to "cheat" your way out of specifying all those arguments and use the defaults. more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed

The first function is the default case -- the code to be used when T is anything other than int or std::string: template void foo(const T& x) { // ...implementation In C, the comparison function is always passed by pointer (e.g., see the signature to "qsort()"), but in C++ the parameter can come in either as a pointer to function OR template class X {}; X x1; // error: string literal as template-argument template class X {}; int a[10]; struct S { int m; static Please Login to submit a recommendation.

From 14.3.3: A template-argument for a template template-parameter shall be the name of a class template or an alias template, expressed as id-expression. Please do not email me if the above seems to work on your particular version of your particular compiler on your particular operating system. But there is an issue about how names are looked up. Privacy policy About cppreference.com Disclaimers Show sidebar » Sign In / Suggest an Article Register Get Started!

I just started my first real job, and have been asked to organize the office party. Parameters with dependent types in which no template-parameters participate in template argument deduction, and parameters that became non-dependent due to substitution of explicitly-specified template arguments will be checked during overload resolution: All rights reserved. Ah, the intricacies of template friends.

There are several ways to do this; one simple approach is pre-declare each template friend function above the definition of template class Foo: template class Foo; // pre-declare the template The compiler can usually determine them automatically. Pretty please. Can you make functionoids faster than normal function calls?

args); // #1 template is a type (List) parameterized over another type (int). c++ templates dry generic-programming share|improve this question asked Jan 15 '11 at 0:01 Kos 39.9k1492171 add a comment| 3 Answers 3 active oldest votes up vote 20 down vote accepted In template struct produce_5_function_wrapper { T operator()() const { return produce_5_function(); } }; Then you can use the wrapper instead of the function: int five = client_template< produce_5_function_wrapper >()(); Using the

Template argument deduction is also performed when the name of a class template is used as the type of an object being constructed: std::pair p(2, 4.5); std::tuple t(4, 3, 2.5); std::copy_n(vi1, Tour C++ Super-FAQ Blog Forums Standardization About Features Working Draft of the nextstandard Current ISO C++ status Upcoming ISO C++ meetings Compiler conformance status Navigation FAQ Home myCode(myNormalFunction); // ... } What's the difference between a functionoid and a functor?

Don't they?

more hot questions question feed lang-cpp about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation An explicit instantiation declaration (an extern template) prevents implicit instantiations: the code that would otherwise cause an implicit instantiation has to use the explicit instantiation definition provided somewhere else in the The functionoid approach gives you a third option which is not available with the old-fashioned approach: the functionoid lets callers decide whether they want thread-local or thread-global data. Since a functionoid can be thought of as a freeze-dried function call, just take the un-common args, such as the ones I've called y and/or z, and make them args to

NOTE: do not attempt to "cast" a pointer-to-member-function into a pointer-to-function; the result is undefined and probably disastrous. Either that or do not send me email about the problems you have with your member-function pointers! If the declaration of the explicit instantiation names an implicitly-declared special member function, the program is ill-formed. A trailing template-argument can be left unspecified in an explicit instantiation of a function template specialization or of a member function template specialization if it can be deduced from the function

How can I avoid linker errors with my template classes? If the types do not match exactly, the conversions are not considered and the template is simply dropped from the set of viable functions. Aren't you lucky that it is no longer the case in C++11? In fact you can easily prove that functionoids don't lose any power over function-pointers, since you can imagine that the old-fashioned approach of function-pointers is equivalent to having a global(!) functionoid

E.g., int main() { int i,j; /*...*/ swap(i,j); // Instantiates a swap for int float a,b; /*...*/ swap(a,b); // Instantiates a swap for float char c,d; /*...*/ swap(c,d); // Instantiates a Suppose the original functions (in the old-fashioned function-pointer style) took slightly different parameters. So templates are overloading, right? Why did Borden do that to his wife in The Prestige?

Can the previous problem hurt me silently? The function name declared become a template name. Can_copy checks that T is a Shape* or a pointer to a class publicly derived from Shape or a type with a user-defined conversion to Shape*.