在 C++ 中干净地将 void 处理为 labmda 的通用返回类型

Question

I am trying to write a function like this:

template <typename T>
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<T(Foo&)> action)
  if (std::is_same<T, void>::value)
  {
    // perform action even if T is void
    action(fooA);
    action(fooB);
  }
  else
  {
    // if T is not void, test also return values
    T resultA = action(fooA);
    T resultB = action(fooB);
    CHECK_EQUAL(resultA, resultB);
  }
  // tests that the state of both foos is the same
  CHECK_EQUAL(fooA, fooB);
}

where T is sometimes void . The compilation (on VS2019) fails with error C2182: 'resultA': illegal use of type 'void' . Is there a clean way around it? (Hopefully one that will compile in most standard compilers)? Thank you.

Answer 1

Right now what's happening is that the compiler still will compile both if branches (since it doesn't actually know which will be called until runtime). This results in a failure since one of the branches doesn't compile correctly. There are a couple fixes (from the comments):

If your compiler supports it this would be one option:

  if constexpr (std::is_same<T, void>::value) {
    // perform action even if T is void
    action(fooA);
    action(fooB);
  } else {
    ...
  }

this will only actually compile one branch depending on the type of T.

If your compiler doesn't support it, here's another option:

template <typename T>
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<T(Foo&)> action) {
    // check assuming T != void
}

// overload which will be called if T was void
void testActionBehavesIdentically(Foo& fooA, Foo& fooB, std::function<void(Foo&)> action) {
    // check assuming T == void
}

The compiler will match the overload when T is void instead of dispatching to the generic version of the function. Here you could even say "if T is an int do something different" which is kind of neat (but can get messy).