function-pointers

I read somewhere that a lambda function should decay to function pointer if the capture list is empty. The only reference I can find now is n3052 . With g++ (4.5 & 4.6) it works as expected, unless the lambda is declared within template code. For example the following code compiles: void foo() { void (*f)(void) = []{}; } But it doesn't compile anymore when templated (if foo is actually called elsewhere): template<class T> void foo() { void (*f)(void) = []{}; } In the reference above, I don't see an explanation of this behaviour. Is this a temporary limitation of g++, and if not, is there a

How can I detect the return type and parameter types of nullary and unary function pointers, std::function objects, and functors (including lambdas)? Boost's function_traits and functional traits don't quite get me there out of the box, but I'm open to supplementing or replacing them. I could do something like this: namespace nsDetail { class Dummy { Dummy(); }; } template<class Fn> struct FnTraits; template<class R> struct FnTraits<R(*)()> { typedef nsDetail::Dummy ParamType; typedef R ReturnType; typedef R Signature(); }; template<class R, class P> struct FnTraits<R(*)(P)> { typedef P

One of my personal experiments to understand some of the C++0x features: I'm trying to pass a function pointer to a template function to execute. Eventually the execution is supposed to happen in a different thread. But with all the different types of functions, I can't get the templates to work. #include <functional> int foo(void) {return 2;} class bar { public: int operator() (void) {return 4;}; int something(int a) {return a;}; }; template <class C> int func(C&& c) { //typedef typename std::result_of< C() >::type result_type; typedef typename std::conditional< std::is_pointer< C >::value,

How (in GCC/"GNU C") do you declare a function pointer which points to an __attribute__((const)) function? The idea being that I want the compiler to avoid generating multiple calls to the function called through the function pointer when it can cache the return value from a previous call. typedef void (*t_const_function)(void) __attribute__((const)); static __attribute__((const)) void A(void) { } static void B(void) { } int main(int argc, const char* argv[]) { t_const_function a = A; // warning: initialization makes qualified // function pointer from unqualified: t_const_function b = B;

Given the following two typedef s: typedef void (*pftype)(int); typedef void ftype(int); I understand that the first defines pftype as a pointer to a function that takes one int parameter and returns nothing, and the second defines ftype as a function type that takes one int parameter and returns nothing. I do not, however, understand what the second might be used for. I can create a function that matches these types: void thefunc(int arg) { cout << "called with " << arg << endl; } and then I can create pointers to this function using each: int main(int argc, char* argv[]) { pftype pointer_one

This question already has answers here : Writing a function pointer in c (3 answers) How do you read C declarations? (10 answers) Closed 4 years ago . Can anyone please explain what int ((*foo(int)))(int) in this does? int (*fooptr)(int); int ((*foo(int)))(int); // Can't understand what this does. int main() { fooptr = foo(0); fooptr(10); } . haccks int ((*foo(int)))(int); This declares foo as a function that expects an int type argument and returns a pointer to a function that expects an int type argument and return an int . To be more clear: foo -- foo foo( ) -- is a function foo(int) --

Following code can't be compiled with g++ version 5.4.0 with option -std=c++1y : void f(int=0) ; int main() { f(); // ok (*f)(2);// ok (*f)();// ok c++11; error with c++14: too few arguments to function return 0; } The function declared to have default argument, so what is wrong here? thanks for help. And why does g++ -c -std=c++11 compile? Accepting (*f)() as valid is a GCC bug. The letter of the standard indicates that using a function name with unary * should cause the function name to decay into a pointer. The pointer should then be dereferenced to obtain the functions address for the call

i basically want to use a dif function to extract a different element of a class (ac). the code is similar to this: .h: class MyClass { public: double f1(AnotherClass &); void MyClass::f0(AnotherClass & ac, double(MyClass::*f1)(AnotherClass &)); }; .cc: double MyClass::f1(AnotherClass & ac) { return ac.value; } void MyClass::f0(AnotherClass & ac, double(MyClass::*f1)(AnotherClass &)) { std::cout << f1(ac); } didn't work, it gives error#547 "nonstandard form for taking the address of a member function" EDIT: I call it from: void MyClass(AnotherClass & ac) { return f0(ac,&f1); // original and