lambda closure type and default argument in function template

lambda closure type and default argument in function template

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lambda closure type and default argument in function template
Tag : cpp , By : Jimmy
Date : November 25 2020, 01:01 AM

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template argument type deduction from std::function return type with lambda

Tag : cpp , By : apple
Date : March 29 2020, 07:55 AM
seems to work fine Don't use std::function in tandem with template argument deduction. In fact, there's very likely no reason to use std::function in a function or function template argument list. More often than not, you should not use std::function; it is a very specialized tool that is very good at solving one particular problem. The rest of the time, you can dispense with it altogether.
In your case you don't need template argument deduction if you use a polymorphic functor to order things:
struct less {
    template<typename T, typename U>
    auto operator()(T&& t, U&& u) const
    -> decltype( std::declval<T>() < std::declval<U>() )
    { return std::forward<T>(t) < std::forward<U>(u); }

    // operator< is not appropriate for pointers however
    // the Standard defines a 'composite pointer type' that
    // would be very helpful here, left as an exercise to implement
    template<typename T, typename U>
    bool operator()(T* t, U* u) const
    { return std::less<typename std::common_type<T*, U*>::type> {}(t, u); }
template<typename Iter, typename Criterion, typename Comparator = less>
void sort_by(Iter first, Iter last, Criterion crit, Comparator comp = less {});

c++ template non-type argument lambda function

Tag : cpp , By : Pat
Date : March 29 2020, 07:55 AM
around this issue Just create a class template with a type parameter, and use decltype to deduce the type of the lambda when you instantiate the template.
#include <functional>

template <typename Function> 
class Bar 
{ };

auto barFunc = [] ( int n ) -> bool { return true; };

int main()
    auto b = Bar<decltype(barFunc)>();
template <typename Function> 
class Bar 

    Bar(Function f) : m_function(f)
    { }


    Function m_function;

How to infer a function type parameter in a template function with a lambda passed as argument?

Tag : cpp , By : Kristian Hofslaeter
Date : March 29 2020, 07:55 AM
I hope this helps . The actual problem is that a lambda function has it own type, that cannot be reduced to R(&)(T). Because of that, C is an incomplete type as correctly outlined by the compiler.
auto p3 = makeC(*(+[](int a){return a;}));
template<typename T>
auto makeC(T&& fun) -> C<decltype(*(+std::forward<T>(fun)))> {
    return C<decltype(*(+std::forward<T>(fun)))>(std::forward<T>(fun));
#include <utility>
#include <string>

template<typename T>
class C: T {
    template<typename F>
    C(F&& fun): T{std::forward<F>(fun)} {}

template<typename R, typename T>
class C<R(&)(T)> {
    template<typename F>
    C(F&& fun) {}

template<typename T>
C<T> makeC(T&& fun) {
    return C<T>(std::forward<T>(fun));

int foo(int a){return a;}

int main() {
    auto p1 = makeC(foo);
    auto p2 = C<int(&)(int)>([](int a){return a;});
    auto p3 = makeC([](int a){return a;});

Pass lambda by argument (no function type template)

Tag : cpp , By : anov
Date : March 29 2020, 07:55 AM
this will help The std::function argument is not deduced, because lamdbas are distinct types convertible to a std::function object, but this conversion is not part of type deduction. Here is how it should work with std::function:
template<typename T>
void mapll(std::function<T(T)> f, std::list<T>& l) {
    for(auto it = l.begin(); it != l.end(); ++it) {
        *it = f(*it);

/* ... */

mapll(std::function<int(int)>{[](int n) { return n * 2; }}, l);
template<typename T>
void mapll(int (*f)(int),  std::list<T>& l) {
    /* Function body identical to the one above. */

/* ... */

/* The stateless lambda converts to the function pointer: */
mapll([](int n) { return n * 2; }, l);
template<typename T>
void mapll(T (*f)(T),  std::list<T>& l)
mapll(static_cast<int(*)(int)>([](int n) { return n * 2; }), l);

What can be said about type of lambda closure defined in function template?

Tag : cpp , By : Ben Kohn
Date : March 29 2020, 07:55 AM
I wish this help you It's true that the standard doesn't say exactly what "unique" means, but I think we can figure it out.
First, notice that the standard says "unique, unnamed non-union class type". The adjective "unnamed" is strong enough to imply that a lambda closure type is never the same as some named class defined using the struct or class keyword or by the standard library (for example, a lambda closure type cannot be std::function<...>).
auto f1 = [](int x) { return x; };
auto f2 = [](int x) { return 2*x; };
auto f3 = [](int x) { return x; };
auto f4 = [](int x) { return x; };
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