定义一个新的数据类型,它的行为就像一个标准的c++数据类型。

时间:2022-03-23 15:53:14

In my program I want to define a new data-type for IPAddressV4 for this purpose I defined the following class:

在我的程序中,我想为IPAddressV4定义一个新的数据类型,为此我定义了以下类:

#include <boost/array.hpp>
#include <boost/asio.hpp>

#include "archive.hpp"

using boost::asio::ip::address_v4;
typedef boost::uint8_t     uint8;   /**< 8-bit unsigned interger.   */

class IPAddrV4
{
public:
/**
 * Construct a IPAddrV4 data type.
 */
IPAddrV4() : _ip_address(0)
{}


/**
 * Construct a IPAddrV4 data type.
 *
 * @param ip_address The value of the IPv4 Address in text format.
 */
IPAddrV4(std::string ip_address) : _ip_address(0)
{
    *this = ip_address;
}

/**
 * Set the IPv4 Address value.
 *
 * @param val The value of the IPv4 Address in text format.
 * @return The reference to the IPAddrV4 data type.
 */
IPAddrV4& operator=(std::string ip_address)
{
    boost::asio::ip::address_v4::bytes_type bytes
            = boost::asio::ip::address_v4::from_string(ip_address).to_bytes();

    _ip_address = ((uint32(bytes.at(0)) << 24) |
                   (uint32(bytes.at(1)) << 16) |
                   (uint32(bytes.at(2)) <<  8) |
                    uint32(bytes.at(3)));

    return *this;
}

IPAddrV4& operator=(uint32 ip_address)
{
    _ip_address = ip_address;
    return *this;
}

/**
 * Get the IPv4 Address data type value.
 *
 * @return The IPv4 Address data type value in binary form.
 */
operator uint32()
{
    return _ip_address;
}

private:
    uint32 _ip_address;

};

Later I am using this class to do some serialization and deserialization to send data over a TCP connection. When I am doing serialization with an output archive, I don't have any problem.

稍后我将使用这个类来执行一些序列化和反序列化以通过TCP连接发送数据。当我使用输出存档进行序列化时,我没有任何问题。

IPAddrV4 ip("127.0.0.1");
oarchive & ip;

In the above example, the value of the IP is converted into binary form and then copied to the output archive. And the oarchive will increase its size by 4 Bytes.

在上面的示例中,IP的值转换为二进制形式,然后复制到输出存档。oarchive将会增加4个字节。

But when I try to use this data-type with the input archive for serialization I get problem:

但是当我尝试使用这个数据类型和输入存档进行序列化时,我遇到了问题:

IPAddrV4 ip;
iarchive & ip;

This is the error which I get while compiling:

这是我在编译时得到的错误:

No match for 'operator&' in 'input_Archive & IPAddrV4::operator uint32()'

So what am I missing in the IPAddrV4?

那么IPAddrV4中缺少什么呢?

Note: The input archive works perfectly with the standard c++ data-types like uint,string..etc.

注意:输入存档与标准c++数据类型(如uint、string..等)完全一致。

2 个解决方案

#1


1  

Boost Serialization has the BOOST_STRONG_TYPEDEF for this purpose.

为此,Boost序列化具有BOOST_STRONG_TYPEDEF。

See also

另请参阅

#2


1  

I found the error, I have to change the overloading of the operator uint32() to be uint32&() as following:

我发现了错误,我必须更改操作符uint32()的重载为uint32&()如下:

operator uint32 &()
{
    return _ip_address;
}

In this way I return value by reference not by value for the input archive.

通过这种方式,我返回的值是引用而不是输入存档的值。

#1


1  

Boost Serialization has the BOOST_STRONG_TYPEDEF for this purpose.

为此,Boost序列化具有BOOST_STRONG_TYPEDEF。

See also

另请参阅

#2


1  

I found the error, I have to change the overloading of the operator uint32() to be uint32&() as following:

我发现了错误,我必须更改操作符uint32()的重载为uint32&()如下:

operator uint32 &()
{
    return _ip_address;
}

In this way I return value by reference not by value for the input archive.

通过这种方式,我返回的值是引用而不是输入存档的值。