模板类型检查参数C ++，不同类型的执行路径

Question

I'd like slightly different logic in a template based on the template parameter. How might I type check a template parameter?

I have the following which I am surprised does not work:

class Bar {
  Bar() {}
}

template <typename T>
void Foo(const &T a) {
  if (!std::is_same<T, Bar>::value) {
    // Do things that will fail for type Bar e.g.
    a.Fail();
  }
}

I do not wish to use template specialization as in reality template specialization ends up sharing a lot of code for my specific purpose (currently working code is using template specialization)

Currently this fails during compile: "no member "Fail" in "Bar"

Answer 1

Rather than specializing the entire template function Foo , specialize a helper method:

template <typename T>
void FooHelper(const T &a) {
    a.fail();
}

template <>
void FooHelper<Bar>(const Bar &a) {
    // something other than a.fail()
}

template <typename T>
void Foo(const T &a) {
    FooHelper<T>(a);
    // etc. etc.
}

Answer 2

Each branch should be valid for every type.
In C++17, you could use if constexpr which change that:

template <typename T>
void Foo(const &T a) {
  if constexpr (!std::is_same<T, Bar>::value) {
    // Do things that will fail for type Bar e.g.
    a.Fail();
  }
}

Before, you have to rely on specialization or overload. As for example

template <typename T>
void Fail(const T& t) { t.Fail(); }

void Fail(const Bar&) { /*Empty*/ }

template <typename T>
void Foo(const &T a) {
    // ...
    Fail(a);
    // ...
}