我知道如何在初始化器列表中初始化const成员,但这需要知道调用构造函数时已经分配的值。据我了解,在Java中,可以在构造函数主体中初始化最终成员,但是我在C (Java的最终与C 的const(中没有看到等效的
但是,当初始化相对复杂时该怎么办?我能提出的最好的方法是具有直接返回实例的初始化函数。还有一些简洁的东西吗?
这是一个示例(https://ideone.com/txxiho(
class Multiplier
{
const int mFactor1;
const int mFactor2;
static void initializationLogic (int & a, int & b, const int c )
{
a = c * 5;
b = a * 2;
}
public:
Multiplier (const int & value1, const int & value2)
: mFactor1(value1), mFactor2(value2)
{};
/*
//this constructor doesn't initialize the const members
Multiplier (const int & value)
{
initializationLogic(mFactor1,mFactor2, value);
};
*/
//this initializes the const members, but it's not a constructor
static Multiplier getMultiplierInstance (const int & value)
{
int f1, f2;
initializationLogic(f1,f2, value);
Multiplier obj(f1,f2);
return obj;
}
您可以将对象构造委派给专用的构造函数,该构造函数接受辅助类的实例,准备所有必要的参数:
private: class
Init_Helper final
{
private: int m_value1;
private: int m_value2;
public: explicit Init_Helper(const int c)
{
m_value1 = c * 5;
m_value2 = m_value1 * 2;
// more init logic goes here...
}
public: int const & Get_Value1(void) const
{
return m_value1;
}
public: int const & Get_Value2(void) const
{
return m_value2;
}
};
public: explicit Multiplier(const int c)
: Multiplier{Init_Helper{c}}
{}
private: explicit Multiplier(Init_Helper && helper)
: mFactor1{helper.Get_Value1()}, mFactor2{helper.Get_Value2()}
{}