Are these not different types? One is a member function pointer, which
requires two arguments - a Real* and an int - whereas the other is a
function pointer which requires one argument - an int;

You can do the following:

template<typename C, typename F>
auto user(C &c, F func){
    if constexpr(std::is_member_function_pointer<F>::value){
        return (c.*func)(5);
    } else{
        return func(5);
    }
}
And subsequently call via:

*user(real, &Real::inc);*

*user(real, Mock::inc);*

On Mon, 25 May 2020, 1:16 pm Nikolay Mihaylov via Std-Proposals, <
std-proposals_at_[hidden]> wrote:

> Hello,
>
> suppose you have following clases:
>
> struct Mock{
>> static int inc(int){
>> return 0;
>> } static int dec(int){
>> return 0;
>> }
>> };struct Real{
>> Real(int v) : v(v){} int inc(int a) const{
>> return a + v;
>> } int dec(int a) const{
>> return a - v;
>> }private:
>> int v;
>> };template<typename C, typename F>
>> auto user(C &c, F func){
>> return (c.*func)(5);
>> }
>
> This will allow to call:
>
>
> *user(real, &Real::inc);*
> but will not allow to do
>
> *user(real, &Mock::inc);*
>
> However because of generic context, both calls must be possible.
>
> Here is simple implementation of the feature:
>
> #include <type_traits>
>>
>> namespace class_invoke_impl_{
>> template <class T, class F, class... Args>
>> constexpr auto class_invoke_(T &&cl, F func, std::true_type,
>> Args&&... args){
>> return (std::forward<T>(cl).*func)(std::forward<Args>(args)...);
>> }
>
>
>
> template <class T, class F, class... Args>
>> constexpr auto class_invoke_(T const &, F func, std::false_type,
>> Args&&... args){
>> return func(std::forward<Args>(args)...);
>> }
>> }
>
>
>
> template <class T, class F, class... Args>
>> constexpr auto class_invoke(T &&cl, F func, Args&&... args){
>> using namespace class_invoke_impl_;
>> return class_invoke_(std::forward<T>(cl), func,
>> std::is_member_pointer<F>{}, std::forward<Args>(args)...);
>> }
>
>
>
>
> --
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>