Using std::apply on class method-CodePudding

I'm trying to get the following to compile (g -11.2, C 20), but I get:

error: no matching function for call to '__invoke(std::_Mem_fn<void (Foo::*)(int, double)>, std::__tuple_element_t<0, std::tuple<int, double> >, std::__tuple_element_t<1, std::tuple<int, double> >)'
 1843 |       return std::__invoke(std::forward<_Fn>(__f),

Code:

#include <iostream>
#include <tuple>

struct Foo
{
    void bar(const int x, const double y) 
    {  
        std::cout << x << " " << y << std::endl;
    }  


    void bar_apply()
    {  
        // fails
        std::apply(std::mem_fn(&Foo::bar), std::tuple<int, double>(1, 5.0));
    }  
};


int main()
{
    Foo foo;
    foo.bar_apply();
};

CodePudding user response：

As others have pointed out, you need an instance to call the member function. This you can deliver to std::apply by wrapping the function call to a lambda expression or via tuple arguments.

However, I would suggest std::invoke, which is more apt to this task. By this way, no instance wrapping is required.

#include <functional> // std::invoke

std::invoke(&Foo::bar, this, 1, 5.0);

CodePudding user response：

I recommend using C 20 bind_front, which is more lightweight and intuitive. Just like its name, member functions require a specific class object to invoke, so you need to bind this pointer to Foo::bar.

void bar_apply()
{  
  std::apply(std::bind_front(&Foo::bar, this), std::tuple<int, double>(1, 5.0));
}

Demo.

CodePudding user response：

std::mem_fn(&Foo::bar) does not store the instance of Foo like a capturing lambda would do:

std::apply([this](int x, double y) { return bar(x,y); },
           std::tuple<int, double>(1, 5.0));

You therefore need to supply the instance when calling the mem_fn object:

std::apply(std::mem_fn(&Foo::bar), std::tuple<Foo*, int, double>(this, 1, 5.0));

or simpler, using make_tuple:

std::apply(std::mem_fn(&Foo::bar), std::make_tuple(this, 1, 5.0));