모호한 C와 '연산자 ='++ 11 표준 : 이동에 대한 과부하 및 복사 및 스왑 관용구

Question

나는 다음과 같은 오류가 점점 오전 : 다음 코드는 컴파일

[matt ~] g++ -std=c++11 main.cpp -DCOPY_AND_SWAP && ./a.out 
main.cpp: In function ‘int main(int, const char* const*)’: 
main.cpp:101:24: error: ambiguous overload for ‘operator=’ in ‘move = std::move<Test&>((* & copy))’ 
main.cpp:101:24: note: candidates are: 
main.cpp:39:7: note: Test& Test::operator=(Test) 
main.cpp:52:7: note: Test& Test::operator=(Test&&) 

:

#include <iostream> 
#include <unordered_map> 
class Test final { 
public: 
    typedef std::unordered_map<std::string, std::string> Map; 
public: 
    Test(); 
    explicit Test(Map&& map); 
    ~Test(); 
    Test(const Test& other); 
    Test(Test&& test); 
#ifdef COPY_AND_SWAP 
    Test& operator=(Test other); 
#else 
    Test& operator=(const Test& other); 
#endif 
    Test& operator=(Test&& other); 
    size_t Size() const noexcept; 
    friend void swap(Test& lhs, Test& rhs); 
private: 
    friend std::ostream& operator<<(std::ostream& stream, const Test& test); 
private: 
    Map map_; 
}; 
Test::Test() : map_() { 
    std::cerr << "Default constructor called" << std::endl; 
}; 
Test::Test(const Test& other) : map_(other.map_) { 
    std::cerr << "Copy constructor called" << std::endl; 
}; 
Test::Test(Test&& other) : map_(std::move(other.map_)) { 
    std::cerr << "Move constructor called" << std::endl; 
}; 
Test::Test(Map&& map) : map_(std::move(map)) { 
    std::cerr << "Map constructor called" << std::endl; 
}; 
Test::~Test() {}; 
#ifdef COPY_AND_SWAP 
Test& Test::operator=(Test other) { 
    std::cerr << "Copy and swap assignment called" << std::endl; 
    using std::swap; 
    swap(this->map_, other.map_); 
    return *this; 
} 
#else 
Test& Test::operator=(const Test& other) { 
    std::cerr << "Copy assignment called" << std::endl; 
    this->map_ = other.map_; 
    return *this; 
} 
#endif 
Test& Test::operator=(Test&& other) { 
    std::cerr << "Move assignment called" << std::endl; 
    this->map_ = other.map_; 
    other.map_.clear(); 
    return *this; 
} 
size_t Test::Size() const noexcept { 
    return map_.size(); 
} 
void swap(Test& lhs, Test& rhs) { 
    using std::swap; 
    swap(lhs.map_, rhs.map_); 
} 
std::ostream& operator<<(std::ostream& stream, const Test& test) { 
    return stream << test.map_.size(); 
} 
int main (const int argc, const char * const * const argv) { 
    using std::swap; 
    Test::Map map { 
    {"some", "dummy"}, 
    {"data", "to"}, 
    {"fill", "up"}, 
    {"the", "map"} 
    }; 
    std::cout << " map size(): " << map.size() << std::endl; 
    std::cout << "Constructing" << std::endl; 
    Test test(std::move(map)); 
    std::cout << " map.size(): " << map.size() << std::endl; 
    std::cout << "test.Size(): " << test.Size() << std::endl; 
    std::cout << "Copy construction" << std::endl; 
    Test copy(test); 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
    std::cout << "Move construction" << std::endl; 
    Test move(std::move(copy)); 
    std::cout << "move.Size(): " << move.Size() << std::endl; 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
    std::cout << "Swapping" << std::endl; 
    swap(move, copy); 
    std::cout << "move.Size(): " << move.Size() << std::endl; 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
    std::cout << "Swapping back" << std::endl; 
    swap(move, copy); 
    std::cout << "move.Size(): " << move.Size() << std::endl; 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
    std::cout << "Copy assignment" << std::endl; 
    copy = test; 
    std::cout << "test.Size(): " << test.Size() << std::endl; 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
    std::cout << "Move assignment" << std::endl; 
    move = std::move(copy); 
    std::cout << "move.Size(): " << move.Size() << std::endl; 
    std::cout << "copy.Size(): " << copy.Size() << std::endl; 
return 0; 
} 

을

는 g++ -std=c++11 main.cpp && ./a.out 컴파일 할 때 :

이

[matt ~] g++ -std=c++11 main.cpp && ./a.out 
map size(): 4 
Constructing 
Map constructor called 
map.size(): 0 
test.Size(): 4 
Copy construction 
Copy constructor called 
copy.Size(): 4 
Move construction 
Move constructor called 
move.Size(): 4 
copy.Size(): 0 
Swapping 
move.Size(): 0 
copy.Size(): 4 
Swapping back 
move.Size(): 4 
copy.Size(): 0 
Copy assignment 
Copy assignment called 
test.Size(): 4 
copy.Size(): 4 
Move assignment 
Move assignment called 
move.Size(): 4 
copy.Size(): 0 

누군가가 모호성이를 사용할 때 발생하는 이유를 내가 이해하는 데 도움이 수 이 경우에 복사 및 교환 관용구? 변환 할 내재적 변환 시퀀스 테스트하고 테스트 & &하기 때문에 과부하 해상도 상업적

Answer 1

기능

Test& operator=(Test other); 
Test& operator=(Test&& other); 

가 동일한 각각 동일하다. 직접 참조 바인딩은 ID 변환으로 간주되기 때문에 전자는 더 좋지 않습니다.

2 개의 똑같이 좋은 일치가 모호한 경우 컴파일러에서 오류가 발생합니다. 당신은 아마도 이것을 원할 것입니다 :

#ifdef COPY_AND_SWAP 
    Test& operator=(Test other); 
#else 
    Test& operator=(const Test& other); 
    Test& operator=(Test&& other); 
#endif