C ++主要是冗余的默認和參數化構造函數，違反了DRY

Question

（在Visual Studio中使用C ++）

我有以下默認構造函數用於創建一個太空船對象：

Ship() // default constructor
{
    name = "[ship unnamed]";
    length = 1000;
    width = 500;
    power = 100;

    vector<int> temp = { 100, 100 }; // { current health, maximum health}

    bridge = temp;
    sensor_arrays.push_back(temp); // 2 sensor arrays
    sensor_arrays.push_back(temp);
    for (int i = 0; i < 12; i++) // create 12 each
    {
        lasers.push_back(temp);
        heavy_lasers.push_back(temp);
        strike_fighters.push_back(temp);
        strike_bombers.push_back(temp);
    }
}

然后我有以下參數化構造函數用於創建給定名稱的船：

Ship(string custom_name)
{
    name = custom_name;
    length = 1000;
    width = 500;
    power = 100;

    vector<int> temp = { 100, 100 }; // { current health, maximum health}

    bridge = temp;
    sensor_arrays.push_back(temp); // 2 sensor arrays
    sensor_arrays.push_back(temp);
    for (int i = 0; i < 12; i++) // create 12 each
    {
        lasers.push_back(temp);
        heavy_lasers.push_back(temp);
        strike_fighters.push_back(temp);
        strike_bombers.push_back(temp);
    }
}

只有一條線路發生了變化，因此這似乎違反了DRY。

我可以使用默認構造函數然后手動更改我需要的內容，但是我希望有一個或多個參數化構造函數而不重復相同的代碼行。 有辦法做到這一點嗎？

Answer 1

您可以使用以下其中一項：

explicit Ship(const std::string& custom_name = "[ship unnamed]") {/*your code*/}

要么

Ship() : Ship("[ship unnamed]") {} // delegate constructor, require C++11
explicit Ship(const std::string& custom_name) {/*your code*/}

Answer 2

感謝@克里斯的回答，我實現描述的初始化列表在這里 。

解決方案:(原始代碼有問題）

Ship(string custom_name) : Ship()
{
    name = custom_name;
    /*length = 1000;
    width = 500;
    power = 100;

    vector<int> temp = { 100, 100 };

    bridge = temp;
    sensor_arrays.push_back(temp); // 2 sensor arrays
    sensor_arrays.push_back(temp);
    for (int i = 0; i < 12; i++) // create 12 each
    {
        lasers.push_back(temp);
        heavy_lasers.push_back(temp);
        strike_fighters.push_back(temp);
        strike_bombers.push_back(temp);
    }*/
}