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C++: Disable implicit conversion in specific contexts only

You get one implicit conversion, so burn it with a wrapper

Consider the following class template in C++:

template<class T>
struct A {
  template<class U>
  A(const A<U>& other);
};

This enables conversion between objects of type A<T> and A<U> when T and U are different. Perhaps A is a container class, for example, and we wish to enable conversion between A<int> and A<float> among other things.

The following will compile:

A<float> f();
A<int> i();
A<float> g() {
  return i();
}

While i() returns an A<int>, it can be implicitly converted to an A<float>, and so this is allowed.

But perhaps conversion between these types is expensive, and we want to ensure that the programmer really wants to do so before it is triggered. We might disallow implicit conversion by declaring the A() constructor to be explicit:

template<class T>
struct A {
  template<class U>
  explicit A(const A<U>& other);
};

The above code will now fail to compile with a message such as the following:

> g++ -c test.cpp
test.cpp: In function ‘A<float> g()’:
test.cpp: error: could not convert ‘i()’ from ‘A<int>’ to ‘A<float>’
   11 |   return i();
      |          ~^~
      |           |
      |           A<int>

We could make this work again by using explicit conversion, i.e. replacing return i() with return A<float>(i()). The explicit constructor permits such explicit conversion, but forbids the previous implicit conversion.

But using an explicit constructor affects all usage of A. Perhaps we want implicit conversion to be allowed in general, but prevented in some particular contexts, or to ensure that some particular piece of code avoids a copy. For this, we will need to remove the explicit again, but can create a sort of wrapper type utility:

template<class T>
struct nocopy {
  const T& ref;
  nocopy(const T& ref) : ref(ref) {
    //
  }
  operator const T&() const {
    return ref;
  }
};

This takes (and stores) a reference to an object of some other type. It also has a user-defined conversion function (also known as a cast operator) that returns the original reference it is given.

Now we see the following:

A<float> f();
A<int> i();
A<float> g() {
  return nocopy(f());  // works
}
A<float> h() {
  return nocopy(i());  // does not work!
}
A<float> j() {
  return f();  // works!
}
A<float> k() {
  return i();  // works!
}

> g++ -c test.cpp
test.cpp: In function ‘A<float> h()’:
test.cpp: error: could not convert ‘nocopy<A<int> >(i())’ from ‘nocopy<A<int> >’ to ‘A<float>’
   25 |   return nocopy(i());  // does not work!
      |          ^~~~~~~~~~~
      |          |
      |          nocopy<A<int> >

That is, whenever nocopy is used we effectively disable implicit conversion; this is why h() fails to compile but all the other functions succeed. We might use nocopy as a sort of static assertion to ensure correctness whenever we have implemented a function that should not result in a copy on return.

This approach works by relying on the rules for implicit conversion (quoting, emphasis mine):

Implicit conversion sequence consists of the following, in this order:

1. zero or one standard conversion sequence;
2. zero or one user-defined conversion;
3. zero or one standard conversion sequence (only if a user-defined conversion is used).

The one user-defined conversion allowed is burned by the nocopy wrapper, so that the user-defined conversion operator in A cannot be used.

Complete code for further play:

template<class T>
struct A {
  template<class U>
  A(const A<U>& other);
};

template<class T>
struct nocopy {
  const T& ref;
  nocopy(const T& ref) : ref(ref) {
    //
  }
  operator const T&() const {
    return ref;
  }
};

A<float> f();
A<int> i();

A<float> g() {
  return nocopy(f());  // works
}
A<float> h() {
  return nocopy(i());  // does not work!
}
A<float> j() {
  return f();  // works!
}
A<float> k() {
  return i();  // works!
}