如果可能enable_if_t如何将模板限制应用于具有单独实现代码的整个类？

问：

注意：我更喜欢始终将我的实现与我的声明分开，即使对于仍必须放在标头中的模板代码也是如此。因此，我倾向于为声明提供一个文件，并在该文件的底部包含一个用于内联和模板实现的文件，然后将我的非内联、非模板实现放在文件中。在这种情况下，可能无法遵循这种偏好，或者我可能做错了什么。.h.hpp.cpp

我有这个初始代码，它编译并运行：

#include <iostream>
#include <memory>
#include <string>

#define OddSource_Export __attribute((visibility("default")))

class IPAddress {};
class IPv4Address : public IPAddress {
public:
    uint8_t version() const {
        return 4;
    }
};
class IPv6Address : public IPAddress {
public:
    uint8_t version() const {
        return 6;
    }
};

template<class TAddress>
class OddSource_Export InterfaceIPAddress
{
public:
    InterfaceIPAddress(TAddress const &, uint16_t);
    uint8_t version() const;
private:
    ::std::unique_ptr<TAddress> const _address;
    uint16_t const _flags;
};

template<class TAddress>
InterfaceIPAddress<TAddress>::
InterfaceIPAddress(TAddress const &address, uint16_t flags)
    : _address(new TAddress(address)),
      _flags(flags)
{}

template<class TAddress>
uint8_t
InterfaceIPAddress<TAddress>::
version() const
{
    return this->_address->version();
}

int main()
{
    IPv4Address addr4;
    InterfaceIPAddress<IPv4Address> address(addr4, 0);
    std::cout << "Version: " << std::to_string(address.version()) << std::endl;
    return 0;
}

但是我想要一个简单的基类限制。因此，我首先定义了这个别名：

template<class TAddress>
using Enable_If_IPAddress = ::std::enable_if_t<::std::is_base_of_v<IPAddress, TAddress>>;

更改我的类声明，如下所示：

template<class TAddress, typename = Enable_If_IPAddress<TAddress>>
class OddSource_Export InterfaceIPAddress

并更改我的构造和方法实现如下：

template<class TAddress, typename>

这是新版本：

#include <iostream>
#include <memory>
#include <string>
#include <type_traits>

#define OddSource_Export __attribute((visibility("default")))

class IPAddress {};
class IPv4Address : public IPAddress {
public:
    uint8_t version() const {
        return 4;
    }
};
class IPv6Address : public IPAddress {
public:
    uint8_t version() const {
        return 6;
    }
};

template<class TAddress>
using Enable_If_IPAddress = ::std::enable_if_t<::std::is_base_of_v<IPAddress, TAddress>>;

template<class TAddress, typename = Enable_If_IPAddress<TAddress>>
class OddSource_Export InterfaceIPAddress
{
public:
    InterfaceIPAddress(TAddress const &, uint16_t);
    uint8_t version() const;
private:
    ::std::unique_ptr<TAddress> const _address;
    uint16_t const _flags;
};

template<class TAddress, typename>
InterfaceIPAddress<TAddress>::
InterfaceIPAddress(TAddress const &address, uint16_t flags)
    : _address(new TAddress(address)),
      _flags(flags)
{}

template<class TAddress, typename>
uint8_t
InterfaceIPAddress<TAddress>::
version() const
{
    return this->_address->version();
}

int main()
{
    IPv4Address addr4;
    InterfaceIPAddress<IPv4Address> address(addr4, 0);
    std::cout << "Version: " << std::to_string(address.version()) << std::endl;
    return 0;
}

但是，现在我的构造函数无法编译：

warning: missing 'typename' prior to dependent type name InterfaceIPAddress<TAddress>::InterfaceIPAddress; implicit 'typename' is a C++20 extension [-Wc++20-extensions]
    InterfaceIPAddress<TAddress>::
    ^
    typename 
error: expected ')'
    InterfaceIPAddress(TAddress const &address, uint16_t flags)
                                ^
note: to match this '('
    InterfaceIPAddress(TAddress const &address, uint16_t flags)
                      ^
error: expected ';' at end of declaration
        : _address(new TAddress(address)),
        ^
        ;
error: no template named '_address'; did you mean 'TAddress'?
        : _address(new TAddress(address)),
          ^~~~~~~~
          TAddress

该方法失败并出现类似的错误。这是在 Clang14 上。GCC 也有各种编译器错误，但它们是不同的：version()

error: invalid use of incomplete type ‘class InterfaceIPAddress’
   38 | InterfaceIPAddress(TAddress const &address, uint16_t flags)
      |                                                           ^
main.cpp:26:24: note: declaration of ‘class InterfaceIPAddress’
   26 | class OddSource_Export InterfaceIPAddress
      |                        ^~~~~~~~~~~~~~~~~~

我很确定我定义正确，因为如果我移动我的实现以与声明内联，它会在没有警告或错误的情况下编译。所以这有效：enable_if_t

#include <iostream>
#include <memory>
#include <string>
#include <type_traits>

#define OddSource_Export __attribute((visibility("default")))

class IPAddress {};
class IPv4Address : public IPAddress {
public:
    uint8_t version() const {
        return 4;
    }
};
class IPv6Address : public IPAddress {
public:
    uint8_t version() const {
        return 6;
    }
};

template<class TAddress>
using Enable_If_IPAddress = ::std::enable_if_t<::std::is_base_of_v<IPAddress, TAddress>>;

template<class TAddress, typename = Enable_If_IPAddress<TAddress>>
class OddSource_Export InterfaceIPAddress
{
public:
    InterfaceIPAddress(TAddress const & address, uint16_t flags)
        : _address(new TAddress(address)),
          _flags(flags)
    {}
    uint8_t version() const
    {
        return this->_address->version();
    }
private:
    ::std::unique_ptr<TAddress> const _address;
    uint16_t const _flags;
};

int main()
{
    IPv4Address addr4;
    InterfaceIPAddress<IPv4Address> address(addr4, 0);
    std::cout << "Version: " << std::to_string(address.version()) << std::endl;
    return 0;
}

那么，我是否遗漏了一些需要为分离实现执行的额外步骤？还是在使用时无法分离此实现？enable_if_t

此外，我尝试在编译器建议的地方添加一个，但所做的只是满足警告。未解决编译错误。typename

编辑：这应该重新打开

我不明白为什么这个问题被关闭了。它说，“编辑问题以包括所需的行为、特定问题或错误，以及重现问题所需的最短代码。这些编辑已经完成，这个问题不可能比现在更清楚了。另外，它已经有一个完整且公认的答案。关闭这个问题是荒谬的，并且阻碍了其他人找到有用信息的能力。