Copy constructor and overloaded addition operator

Question

I am reviewing operator overloading in C++. Just for fun I am implementing a BigInt class.

The first operator I want to overload for it is the addition operator. I have decided to overload this operator as a friend non-member function. Here's a MWE of this code:

#include <cassert>
#include <iostream>
#include <string>

class BigInt{
 public:
  friend BigInt operator+(const BigInt &bi1, const BigInt &bi2);

  BigInt() {}
  explicit BigInt(const std::string &in) {
    if (in.size() != 0) {
      for (auto cc = in.rbegin(); cc != in.rend(); ++cc) {
        value_.push_back(*cc);
      }
    }
  }
  std::string value() {
    std::string actual_value{};  // Reversed string.
    for (auto cc = value_.rbegin(); cc != value_.rend(); ++cc) {
      actual_value.push_back(*cc);
    }
    return actual_value;
  }

 private:
  std::string value_;  // String of digits as characters.
};

BigInt operator+(const BigInt &bi1, const BigInt &bi2) {
  BigInt result{};

  result.value_ = "4421";
  return result;
}

int main() {
  std::cout << "Test addition operator... ";
  std::string number{"1234"};  // Number 1,234.
  BigInt mm(number);
  std::string number_ten{"10"};  // Number 10.
  BigInt nn(number_ten);

  BigInt mm_nn = mm + nn;

  std::string expected_result{"1244"};  // 1,234 + 10 = 1,244.
  assert(mm_nn.value() == expected_result);
  std::cout << "ok." << std::endl;
}

This code mocks the behavior of the addition. It compiles and runs. Yet when I add a copy constructor for the BigInt class, this codes stops working. I.e. if I add this to the class declaration:

explicit BigInt(const BigInt &in): value_(in.value_) {}

The code does not even compile. The addition function as coded returns a copy of a constructed instance of BigInt. For this a copy constructor must be defined. If I do not define it myself, then the compiler will do so. What does the compiler produce that I am not producing with the added copy constructor? Here's the compilation error I get:

$ g++ -std=c++14 -g mwe.cpp 
mwe.cpp: In function ‘BigInt operator+(const BigInt&, const BigInt&)’:
mwe.cpp:34:10: error: no matching function for call to ‘BigInt::BigInt(BigInt&)’
   return result;
          ^
mwe.cpp:9:3: note: candidate: BigInt::BigInt()
   BigInt() {}
   ^
mwe.cpp:9:3: note:   candidate expects 0 arguments, 1 provided
mwe.cpp: In function ‘int main()’:
mwe.cpp:44:23: error: no matching function for call to ‘BigInt::BigInt(BigInt)’
   BigInt mm_nn = mm + nn;
                       ^
mwe.cpp:9:3: note: candidate: BigInt::BigInt()
   BigInt() {}
   ^
mwe.cpp:9:3: note:   candidate expects 0 arguments, 1 provided

From it, it seems like the compiler expects a copy constructor that I have not provided. Now... IF I REMOVE the explicit keyword, everything works. However, I have seen implementations with explicit copy constructor, for example: Explicit copy constructor

What am I missing? Why can't I make this copy constructor explicit while overloading the addition operator? Should, in general, copy constructors be made explicit?

Answer 1

Making the copy constructor explicit does not make sense. Remove it.

BigInt(const BigInt &in): value_(in.value_) {}

---

Conceptual Problem With explicit Copy Constructor

Making a copy constructor explicit makes returning an object from a function impossible.

Let's simplify your code to the following:

struct BigInt
{
   BigInt() {}
   explicit BigInt(const BigInt &in) {}
};

BigInt operator+(const BigInt &bi1, const BigInt &bi2)
{
   BigInt result;
   return result;
}

int main() {}

In the line return result, the compiler relies on the copy constructor to return an object. When the copy constructor is explicit, there is no way for a BigInt to be constructed as the return value.

Trying to use:

BigInt operator+(const BigInt &bi1, const BigInt &bi2)
{
   BigInt result;
   return BigInt(result);
}

is futile since that is equivalent to:

BigInt operator+(const BigInt &bi1, const BigInt &bi2)
{
   BigInt result;
   BigInt result1(result);
   return result1;
}

The problem continues to be there no matter what you do in the function.