我有一个(对我而言)复杂对象，它包含大约 20 个数据成员，其中许多是指向其他类的指针。所以对于构造函数，我有一个又长又复杂的初始化列表。该类还具有十几个不同的构造函数，反射(reflect)了可以创建该类的各种方式。这些初始化项目中的大多数在这些不同的构造函数之间都没有变化。

我担心的是，我现在有大量复制(或大部分复制)代码，如果我需要向类中添加新成员，可能无法将其放入每个构造函数初始化列表中。

class Object 
{
    Object();
    Object(const string &Name);
    Object (const string &Name, const string &path);
    Object (const string &Name, const bool loadMetadata);
    Object (const string &Name, const string &path, const bool loadMetadata);
} 

Object::Object() :
    name(),
    parent_index (0),
    rowData (new MemoryRow()),
    objectFile (),
    rows (new MemoryColumn (object_constants::RowName, OBJECTID, object_constants::ROWS_OID)),
    cols (new MemoryColumn (object_constants::ColName, OBJECTID, object_constants::COLS_OID)),
    objectName (new MemoryColumn(object_constants::ObjName, STRING, object_constants::short_name_len, object_constants::OBJECTNAME_OID)),
    parent     (new MemoryColumn(object_constants::ParentName, STRING, object_constants::long_name_len, object_constants::PARENT_OID)),
    parentIndex (new MemoryColumn(object_constants::ParentIndex, OBJECTID, object_constants::PARENTINDEX_OID)),
    childCount (new MemoryColumn (object_constants::ChildCount, INTEGER, object_constants::CHILD_COUNT_OID)),
    childList (new MemoryColumn (object_constants::ChildList, STRING, object_constants::long_name_len, object_constants::CHILD_OID)),
    columnNames (new MemoryColumn (object_constants::ColumnNames, STRING, object_constats::short_name_len, object_constants::COLUMN_NAME)),
    columnTypes (new MemoryColumn (object_constants::ColumnTypes, INTEGER, object_constants::COLUMN_TYPE)),
    columnSizes (new MemoryColumn (object_constants::ColumnSizes, INTEGER, object_constants::COLUMN_SIZE))
{}

然后对其他构造函数重复上述操作。有没有为此使用默认构造函数的聪明方法，然后修改其他构造函数的结果？

最佳答案

如何将公共(public)字段重构为基类。基类的默认构造函数将处理大量默认字段的初始化。看起来像这样:

class BaseClass {
    public:
    BaseClass();
};

class Object : public BaseClass
{
    Object();
    Object(const string &Name);
    Object (const string &Name, const string &path);
    Object (const string &Name, const bool loadMetadata);
    Object (const string &Name, const string &path, const bool loadMetadata);
};

BaseClass::BaseClass() :
    parent_index (0),
    rowData (new MemoryRow()),
    objectFile (),
    rows (new MemoryColumn (object_constants::RowName, OBJECTID, object_constants::ROWS_OID)),
    cols (new MemoryColumn (object_constants::ColName, OBJECTID, object_constants::COLS_OID)),
    objectName (new MemoryColumn(object_constants::ObjName, STRING, object_constants::short_name_len, object_constants::OBJECTNAME_OID)),
    parent     (new MemoryColumn(object_constants::ParentName, STRING, object_constants::long_name_len, object_constants::PARENT_OID)),
    parentIndex (new MemoryColumn(object_constants::ParentIndex, OBJECTID, object_constants::PARENTINDEX_OID)),
    childCount (new MemoryColumn (object_constants::ChildCount, INTEGER, object_constants::CHILD_COUNT_OID)),
    childList (new MemoryColumn (object_constants::ChildList, STRING, object_constants::long_name_len, object_constants::CHILD_OID)),
    columnNames (new MemoryColumn (object_constants::ColumnNames, STRING, object_constats::short_name_len, object_constants::COLUMN_NAME)),
    columnTypes (new MemoryColumn (object_constants::ColumnTypes, INTEGER, object_constants::COLUMN_TYPE)),
    columnSizes (new MemoryColumn (object_constants::ColumnSizes, INTEGER, object_constants::COLUMN_SIZE))
{}

现在，您的 Object 构造函数应该看起来更易于管理:

Object::Object() : BaseClass() {}
Object::Object (const string &Name): BaseClass(), name(Name) {}
Object::Object (const string &Name, const string &path): BaseClass(), name(Name), path_(path){}
Object::Object (const string &Name, const bool loadMetadata): BaseClass(), name(Name){}
Object::Object (const string &Name, const string &path, const bool loadMetadata): BaseClass(), path_(path) {}

本质上与 Iraimbilanja 的回答类似，但避免添加内部类来访问数据，这可能会影响许多现有代码。但是，如果您已经有了类层次结构，则可能很难将其分解为基类。

关于c++ - 处理具有长初始化列表和多个构造函数的类？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/761917/