std-function

I'm playing around with std::function and custom allocators but its not behaving as I expected when I don't provide the function with an initial functor. When I provide a custom allocator to the constructor but no initial functor, the allocator is never used or so it seems. This is my code. //Simple functor class that is big to force allocations struct Functor128 { Functor128() {} char someBytes[128]; void operator()(int something) { cout << "Functor128 Called with value " << something << endl; } }; int main(int argc, char* argv[]) { Allocator<char, 1> myAllocator1; Allocator<char, 2>

I'm having trouble with std::functions created from lambdas if the function returns a reference but the return type isn't explicitly called out as a reference. It seems that the std::function is created fine with no warnings, but upon calling it, a value is returned when a reference is expected, causing things to blow up. Here's a very contrived example: #include <iostream> #include <vector> #include <functional> int main(){ std::vector<int> v; v.push_back(123); std::function<const std::vector<int>&(const std::vector<int>&)> callback = [](const std::vector<int> &in){return in;}; std::cout <<

This question already has an answer here: No viable conversion from std::function to bool 1 answer Consider the following code. #include <functional> int main(void) { std::function<void()> f1; if (f1) { /* ok */ ... } bool b = f1; /* compile-error */ bool B = !f1; /* ok */ ... } std::function<> converts implicitly to bool in some circumstances but not in all of them. Assigning it to a bool -variable does not work, whereas the result of an operation or using it in an if() -statement is OK. Why is that so? It seems we have to do an boolean-operation on it, then the conversion works. What I did

Is it possible to replace one std::function from within itself with another std::function ? The following code does not compile: #include <iostream> #include <functional> int main() { std::function<void()> func = []() { std::cout << "a\n"; *this = std::move([]() { std::cout << "b\n"; }); }; func(); func(); func(); } Can it be modified to compile? The error message right now is: 'this' was not captured for this lambda function - which I completely understand. I don't know, however, how I could capture func 's this -pointer. I guess, it is not even a std::function inside the lambda, yet?! How

After going though a question on std::bind , I was wondering if it was possible to hold a vector of functions created by std::bind so I can avoid using std::function and its heavyweight wrapping. #include <iostream> #include <functional> #include <typeinfo> #include <vector> int add(int a, int b) {return a + b;} int main() { //I believe this here is just a special type of bound function. auto add2 = std::bind(add, std::placeholders::_1, 2); auto add3 = std::bind(add, std::placeholders::_1, 3); //Yup. std::cout << typeid(add2).name() << std::endl; //Here's the type of the second function std:

I have three functions I'm looking to merge together. Each one takes an std::function as the first parameter, then executes it within a try / catch block. The issue is, there are three different types of functions. Functions with no parameters, those with one integer parameter, and those with two integer parameters. The ones with integer parameters also have their corresponding parameters passed through the original function. As one can see, each of the functions are nearly identical, so it would be nice if I could merge them all together. However, I'm unsure of anyway to setup a parameter

Given the type of a callable function C , I want to get at compile time a std::function ; the type of which: has the same return type of function C the argument types are the first N argument types of function C This means that, for a given type void(int, char, double) and a given N , the type of the function is: N = 1 => result type: std::function<void(int)> N = 2 => result type: std::function<void(int, char)> N = 3 => result type: std::function<void(int, char, double)> N > 3 => compile time error Example: template<std::size_t N, typename R, typename... A> constexpr auto get() { return /*