How Can I Prevent Template Generation for Objects That Don't Implement a Method

Question

So for the purpose of example let\'s say I have 3 simple structs, the second of which doesn\'t contain a bar method:

struct one {
          


        

Answer 1

First, the utilities. One is our favorite overload that shows off three C++17 features, and overloads the operator() of several function objects, all in two lines.

template<class... Ts> struct overload : Ts... { using Ts::operator()...; };
template<class... Ts> overload(Ts...) -> overload<Ts...>;

Then an accepts-everything fallback tag type:

struct fallback_t { template<class T> fallback_t(T&&) {} };

Now to the call itself. We make the original generic lambda SFINAE-friendly by putting the value.bar() call into the trailing return type, then overload it with a fallback overload that has undefined behavior if actually invoked (since OP "omit[ted] any explicit definition of behavior"):

void bar(const int key) { 
    findObject(key, overload {
          [&](auto& value) -> decltype(void(value.bar())) { value.bar(); },
          [](fallback_t){ fire_missiles_and_impregnate_cat(); }
    } ); 
}