Chapter 5. Semantics of Construction, Destruction, and Copy
5.1 “無繼承” 情況下的對(duì)象構(gòu)造
1. 普通類型(和C相同)
2. 抽相數(shù)據(jù)類型
3. 為繼承作準(zhǔn)備
5.2 繼承體系下對(duì)象的構(gòu)造
1.? 通用繼承構(gòu)造規(guī)則
???? (1) 在成員初始化列表中的data members初始化操作會(huì)被放進(jìn)constructor的函數(shù)本身，并以members的聲明順序?yàn)轫樞颉?br />???? (2) 如果有一個(gè)member沒有出現(xiàn)在初始化列表中，但它有一個(gè)default constructor，那么default constructor會(huì)被調(diào)用。
???? (3) 如果class object 有virtual table pointer(s)， 它(們)必須被設(shè)定初始值，指向適當(dāng)?shù)膙irtual table(s)。
???? (4) 在那之前，所有的上一層的base class constructors必須被調(diào)用, 以base class的聲明順序?yàn)轫樞?br />???? (5) 在那之前，所有的virtual base class constructors必須被調(diào)用，從左到右，從最深到最淺。
2. 虛擬繼承(virtual Inheritance)
如對(duì)于下面的類:?
?1?
class?Point3d?:?public?virtual?Point
?2?
?{?
?3?
?public:?
?4?
????Point3d(?float?x?=?0.0,?float?y?=?0.0,?float?z?=?0.0?)?
?5?
????????:?Point(?x,?y?),?_z(?z?)?{}?
?6?
????Point3d(?const?Point3d&?rhs?)?
?7?
????????:?Point(?rhs?),?_z(?rhs._z?)?{}?
?8?
????~Point3d();?
?9?
????Point3d&?operator=(?const?Point3d&?);?
10?
11?
????virtual?float?z(){?return?_z;?}?
12?
????//?
?
13?
?protected:?
14?
????float?_z;?
15?
?};?
?
?可能的轉(zhuǎn)換是這樣的:
?1?//?Psuedo?C++?Code:?
?2??//?Constructor?Augmentation?with?Virtual?Base?class?
?3??Point3d*?Point3d::Point3d(?Point3d?*this,?bool?__most_derived,??float?x,?float?y,?float?z?)?
?4??{?
?5???if?(?__most_derived?!=?false?)?
?6????this->Point::Point(?x,?y);?
?7?
?8???this->__vptr__Point3d?=?__vtbl__Point3d;?
?9???this->__vptr__Point3d__Point?=?__vtbl__Point3d__Point;?
10?
11???this->_z?=?rhs._z;?
12???return?this;?
13??}?
?//
?對(duì)于如下的繼承層次:
?class Vertex?? : virtual public Point { ... };
?class Vertex3d : public Point3d, public Vertex { ... };
?class PVertex? : public Vertex3d { ... };
?類Point3d的構(gòu)造可能是:
?1?//?Psuedo?C++?Code:?
?2??//?Constructor?Augmentation?with?Virtual?Base?class?
?3??Point3d*?Point3d::Point3d(?Point3d?*this,?bool?__most_derived,??float?x,?float?y,?float?z?)?
?4??{?
?5???if?(?__most_derived?!=?false?)?
?6????this->Point::Point(?x,?y);?
?7?
?8???this->__vptr__Point3d?=?__vtbl__Point3d;?
?9???this->__vptr__Point3d__Point?=?__vtbl__Point3d__Point;?
10?
11????this->_z?=?rhs._z;?
12???return?this;?
13??}
?//
?對(duì)于Vertex3d的構(gòu)造可能是如下:
?1?//?Psuedo?C++?Code:?
?2??//?Constructor?Augmentation?with?Virtual?Base?class?
?3??Vertex3d*?Vertex3d::Vertex3d(?Vertex3d?*this,?bool?__most_derived,?float?x,?float?y,?float?z?)?
?4??{?
?5???if?(?__most_derived?!=?false?)?
?6????this->Point::Point(?x,?y);?
?7?
?8???//?invoke?immediate?base?classes,?
?9???//?setting?__most_derived?to?false?
10?
11???this->Point3d::Point3d(?false,?x,?y,?z?);?
12???this->Vertex::Vertex(?false,?x,?y?);?
13?
14???//?set?vptrs?
15???//?insert?user?code?
16?
17???return?this;?
18??}?
?
3. vptr初始化語意學(xué)(The Semantics of the vptr Initialization)
???? (1) 構(gòu)造函數(shù)執(zhí)行算法
????????? I.??? 在derived class constructor 中, 所有的"virtual base classes" 及 "上一層base class"的constructors會(huì)被調(diào)用.
?? II.? 上述完成之后, 對(duì)象的vptr(s)會(huì)被初始化, 指向相關(guān)的virtual talbe(s).
?? III. 如果有成員初始化列表的話, 將在constructor體內(nèi)擴(kuò)展開來. 這必須在vptr被設(shè)定之后才能進(jìn)行,以免有一個(gè)virtual member function被調(diào)用
?? IV. 最后, 執(zhí)行程序所提供的代碼.
????
???? (2) 示例:
???? For example, given the following user-defined PVertex constructor:
1?PVertex::PVertex(?float?x,?float?y,?float?z?)?
2?????:?_next(?0?),?Vertex3d(?x,?y,?z?)
3?????,?Point(?x,?y?)?
4??{?
5???if?(?spyOn?)?
6????cerr?<<?"within?Point3d::Point3d()"??<<?"?size:?"?<<?size()?<<?endl;?
7??}?
?
??可能一個(gè)擴(kuò)展如下:
?1?//?Pseudo?C++?Code?
?2??//?expansion?of?PVertex?constructor?
?3??PVertex*?PVertex::PVertex(?Pvertex*?this,??bool?__most_derived,?float?x,?float?y,?float?z?)?
?4??{?
?5???//?conditionally?invoke?the?virtual?base?constructor?
?6???if?(?__most_derived?!=?false?)?
?7????this->Point::Point(?x,?y?);?
?8???//?unconditional?invocation?of?immediate?base?
?9????this->Vertex3d::Vertex3d(?x,?y,?z?);?
10?
11???//?initialize?associated?vptrs?
12?
13???this->__vptr__PVertex?=?__vtbl__PVertex;?
14???this->__vptr__Point__PVertex?=?__vtbl__Point__PVertex;?
15?
16???//?explicit?user?code?
17???if?(?spyOn?)?
18????cerr?<<?"within?Point3d::Point3d()"?
19????<<?"?size:?"?<<?(*this->__vptr__PVertex[?3?].faddr)(this)?
20????<<?endl;?
21?
22???//?return?constructed?object?
23???return?this;?
24??}?
?
5.3 對(duì)象的復(fù)制(Object Copy Sematics)
1. Copy constructor operator 在以下幾種情況下不會(huì)表現(xiàn)出: bitwisecopy
???? (1) class 內(nèi)帶有一個(gè)member object, 而其類有一個(gè)copy constructor operator時(shí)。
???? (2)? 當(dāng)一個(gè) class 的base 有一個(gè)copy assignment operator時(shí)。
???? (3)? 當(dāng)類聲明了任何一個(gè)virtual functions時(shí)。
???? (4)? 當(dāng)class繼承自一個(gè)virtual base class時(shí)
?2. 合成示例
?1??//?Pseudo?C++?Code:?synthesized?copy?assignment?operator?
?2??inline?Point3d&?Point3d::operator=(?Point3d?*const?this,?const?Point3d?&p?)?
?3??{?
?4???//?invoke?the?base?class?instance?
?5???this->Point::operator=(?p?);?
?6?
?7???//?memberwise?copy?the?derived?class?members?
?8???_z?=?p._z;?
?9???return?*this;?
10??}?
?
5.4 對(duì)象的功能
5.5 析構(gòu)語意學(xué)(Semantics of Destruction)
1. 析構(gòu)函數(shù)生成原則：
?如果類沒有定義destructor, 那么只有在class內(nèi)帶的member object(或是class自己的base class)擁有destructor的情況下，編譯器才會(huì)自動(dòng)的合成一個(gè)來。否則，destructor會(huì)被視為不需要，也就不需要被合成(當(dāng)然更不需要被調(diào)用)
2. 析構(gòu)調(diào)用過程
???? (1) destructor的函數(shù)本身首先被執(zhí)行。
???? (2) 如果class擁有member class objects, 而后者擁有destructors, 那么它會(huì)以其聲明順序的相反順序被調(diào)用.
???? (3) 如果object內(nèi)帶一個(gè)vptr， 則現(xiàn)在被重新設(shè)定，指向適當(dāng)?shù)腷ase class的virtual table.
???? (4) 如果任何直接的(上一層)novirtual base classes 擁有destructor，那么它會(huì)以其聲明順序相反的順序調(diào)用
???? (5) 如果有任何的virtual base classes 擁有destructor， 而當(dāng)前討論的這個(gè)class是最末端的class， 那么它們會(huì)以其原來的構(gòu)造順序相反的順序被調(diào)用.
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