How to obtain result type of an class overloaded method via std::invoke_result

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C++

SOLUTION FOUND : per example below :

auto(cfoo::* _cfoo_foo_int_ptr)(int) = &cfoo::foo;
using result_type = invoke_result<decltype(_cfoo_foo_int_ptr), cfoo, int>::type;

greetings kind regards
i am attempting to determine the result type of an overloaded class method via std::invoke_result . compile errors result in attempt shown below . i seek your kind assistance . thank you kindly

#include <iostream>
#include <type_traits>
using namespace std;

struct cfoo
{
	int foo(int) { cout << __FUNCSIG__ << endl; return 0; }
	int foo(int, int) { cout << __FUNCSIG__ << endl; return 0;}
};

int main()
{
	auto(cfoo::* _cfoo_foo_int_ptr)(int) = &cfoo::foo;
	cfoo _cfoo;
	(_cfoo.*_cfoo_foo_int_ptr)(0); // callable !
	using result_type = invoke_result<decltype(_cfoo_foo_int_ptr), int>::type;
	return 0;
}

#ifdef _COMPILE_ERRORS_BELOW_
1>bugyCode.cpp(16, 71) : error C2039 : 'type' : is not a member of 'std::invoke_result<void (__cdecl cfoo::* )(int),int>'
1>bugyCode.cpp(16, 22) : message: see declaration of 'std::invoke_result<void (__cdecl cfoo::* )(int),int>'
1>bugyCode.cpp(16, 71) : error C2061 : syntax error : identifier 'type'#endif
#endif

What I have tried:

attempted various syntasexeseses , read cppreference re/ invoke_result , consulted with open ai chat bot