一、考試內容與筆記

1. What are the three main characteristics of Object-Oriented Programming?

Encapsulation 封裝：Type = Data + Operation。限制使用權限（public, protected, private），有利於debug，讓錯誤限制在某一個區塊。
Inheritance 繼承：重複使用(reuse)程式碼
Polymorphism 多型：同名的函式與一樣的參數（同樣的signature），但有不同版本的函數，使用vitual，可以解決C中Switch問題

2. Define in C++ a pointer that points to the element a [5] [10] of the array "int a [10] [20] [30] [40] " named p.

答案：int (*p)[30][40] = &a[5][10]

int a [10] [20] [30] [40]

item	type	size
a[0][0][0][0]	int	sizeof(int)
a[0][0][0]	Int [40]	sizeof(int)*40
a[0][0]	Int [30][40]	*sizeof(int)3040*
a[0]	Int [20][30][40]	sizeof(int)2030*40

C的內建矩陣其實就是指標，其矩陣名字就是一個指標，而以上為高維矩陣各個的type與size。而a[5][10]的type為int [30][40]，因此宣告int (*p)[30][40] = &a[5][10]，要記得加&，因為a[5][10]是一個element，要使用取址符號取址。

高維陣列，只有一個指標，所以3維陣列，也只有一個指標，故在宣告a[5][10]時，只要宣告單一指標(*p)即可，不用宣告成雙重指標(**p)。

更多陣列與指標關係，請參考［筆記］陣列與指標｜C++

3. Now, assuming that the value of p in Question 2 is z, what would be the values of p + 1 and p - 1?Why?

答案：(a) z+1200*sizeof(int), (b) z-1200*sizeof(int)

因為a[5][10]的size為sizeof(int)*30*40 = 1200*sizeof(int)，要告訴array每個element的type，才能正確跳躍。單一指標的加/減，會加/減n個element，而在記憶體中會加減n*size byte，所以直接印出指標的值（address），會發現相加/減前後相差n*size byte。

＊指標為儲存記憶體位址 (address)的變數

更多指標運算，請參考［筆記］陣列與指標｜C++

4. Again, assuming that the value of p in Question 2 is z, what would be the values of (p + 1) - p and p - (p + 1)? Why?

答案：(a) 1, (b) -1

因為指向同一個array的指標相減，代表兩指標的距離，故回傳值不會是adress，而是距離，所以其單位是element。兩指標做運算時，一定要有一個為正(+)，一個為負(-)，不能直接對指標做乘(*)除(/)，也不能對兩指標做相加。

更多指標運算，請參考［筆記］陣列與指標｜C++

5. Define in C++ a pointer "p" to the array "int a [256] ;" so that p[1] is an alias of a [0], p[2] is an alias of a [1], and so on, all the way up so that p [256] is an alias of a [255].

答案：int *p = a-1 = &a[-1]

這樣可以實作一個index為1開始的array。

更多陣列與指標關係，請參考［筆記］陣列與指標｜C++

6. In C++, which four of the operators have to be defined as a member function when overloaded. Why?

答案：=, [], (), ->

因為回傳要lvalue

return object // 相當於回傳rvalue
return object reference // 這邊又可細分成回傳rvalue reference或lvalue reference

lvalue與rvalue除了左右關係外：

lvalue：有名字的，例如：變數
rvalue：匿名的，例如：值

&& 和 || 這兩個operator如果要overloading必須小心，e1 && e2 && e3 && e4……，原本如果e1為false，判斷就會結束，但如果overloading，此性質就會消失。

<<, >> voerloading時，要宣告成friend，因為我們不能動ostream物件的東西，故只能授權ostream能存取新物件的private成員。且傳參考給operator，也回傳參考(lvalue)，這樣就可以連續輸入/輸出。

7. What would be the output of the following C++ program?

答案：0,4,16,36

目的：

模板超程式設計（英語：Template metaprogramming，縮寫：TMP）是一種超程式設計技術，編譯器使用模板產生暫時性的原始碼，然後再和剩下的原始碼混合並編譯。這些模板的輸出包括編譯時期常數、資料結構以及完整的函式。如此利用模板可以被想成編譯期的執行。

過程：

template <int I = 0, int ...D> struct Helper : Helper<I+ 1, D..., (I+I)*(I+I)> {};
template<> struct Helper<0, <> > : Helper<0 + 1, (0+0) * (0+0)> {};
template<> struct Helper<1, <0> > : Helper<1 + 1, 0, (1+1) * (1+1)> {};
template<> struct Helper<2, <0, 4> > : Helper<2 + 1, 0, 4, (2+2) * (2+2)> {};
template<> struct Helper<3, <0, 4, 16> > : Helper<3 + 1, 0, 4, 16, (3+3) * (3+3)> {};
template<> struct Helper<4, <0, 4, 16, 36> > {static constexpr std::array<int, S> table = {0, 4, 16, 36};};

程式碼來源：WikiPedia. Template metaprogramming

細節：

constexpr：complier time const variable
#15, 16：是一個可變參數的遞迴模板結構helper，template<>傳入兩種參數，一種為一個int I，另一種為int ...I（可變參數），而helper自己呼叫自己，並傳三種引數給自己分別為(1)I+1, (2) D... (3) (I+I)*(I+I)。故每展開一個遞迴，參數就變多一個。
#19~21：是一個特殊化的模板結構helper，用於終止遞迴
… args : packed parameter
args… : unpacked parameter
helper<>::table：呼叫helper中的table，因為helper模板中已有預設的參數，故<>中可為空模板

8. What would be the output of the following C++ program?

#include <iostream>
int main() {
    int a = 10;
    int b = 20;
    int z = (!!a << 3) | (!!b << 4);
    std::cout << z << std::endl;
}

答案：24

首先要分清楚！和～這兩個差別。

! : logical NOT
~ : bitwise NOT
| : bitwise OR
<< : shift left

然後運算子的先後順序為 () → ！ → << → ｜

int a = 10; // 1010(2)
int b= 20; // 10100(2)
!a == false == 0
!!a == true == 1
!!a << 3 == true << 3 == 1 << 3 == 1000(2) == 8
!!b < <4 == 10000(2) == 16
(!!a << 3) | (!!b << 4) == 1000(2) | 10000(2) == 11000(2) == 24

更多運算子優先順序介紹，請參考［筆記］介面與實作、運算子多載、左值右值、參數傳遞、回傳多值｜C++

9. Is the following C++ program correct? If not, show how to correct it.

template<class T>
class Handle {
    T* p;
    public:
    Handle(T* pp) : p{pp} {} // default constructor
    ~Handle() { delete p; } // destructor 
    Handle(Handle&& h) :p{h.p} { h.p=nullptr; } // move constructor
    Handle& operator=(Handle&& h) { delete p; p=h.p; h.p=nullptr; return *this; } // move assignment operator
};

Handle(Handle&& h) :p{h.p} { h.p=nullptr; } // move constructor, 舊的object一定要接上nullptr，因為如果不接上nullptr，新舊參考都指向同一個東西，舊的object會自動呼叫destructor，則舊的object所指向的東西也會一並刪除，此時新的參考所指向的東西就會被free掉。

Handle& operator=(Handle&& h) { delete p; p=h.p; h.p=nullptr; return *this; } // move assignment operator，少檢查if(this != &h)，因為要避免自我賦值。

程式碼來源：Bjarne Stroustrup. 2016. C++11 - the new ISO C++ standard

更多類別介紹，請參考［筆記］類別、特殊函式、內嵌函式、函式物件｜C++

10. Implement the copy constructor and copy assignment operator for the class in Question 9 so that its instances can only be moved, but not copied.

Handle(const Handle&) = delete; // copy constructor, delete 是指no copy
Handle& operator=(const Handle&) = delete; // copy assignment operator, delete 是指no copy

更多類別介紹，請參考［筆記］類別、特殊函式、內嵌函式、函式物件｜C++

三、相關文章

三、考試資訊

學期：109-2
課程：高等物件導向程式設計
學系：國立中山學資訊工程學系碩士班
老師：江明朝

宗平方 TsungSquare

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［筆記］109-2高等物件導向程式設計期中考

一、考試內容與筆記

1. What are the three main characteristics of Object-Oriented Programming?

2. Define in C++ a pointer that points to the element a [5] [10] of the array "int a [10] [20] [30] [40] " named p.

3. Now, assuming that the value of p in Question 2 is z, what would be the values of p + 1 and p - 1?Why?

4. Again, assuming that the value of p in Question 2 is z, what would be the values of (p + 1) - p and p - (p + 1)? Why?

5. Define in C++ a pointer "p" to the array "int a [256] ;" so that p[1] is an alias of a [0], p[2] is an alias of a [1], and so on, all the way up so that p [256] is an alias of a [255].

6. In C++, which four of the operators have to be defined as a member function when overloaded. Why?

7. What would be the output of the following C++ program?

8. What would be the output of the following C++ program?

9. Is the following C++ program correct? If not, show how to correct it.

10. Implement the copy constructor and copy assignment operator for the class in Question 9 so that its instances can only be moved, but not copied.

三、相關文章

三、考試資訊

四、參考資料

留言

張貼留言

［筆記］109-2高等物件導向程式設計 期中考

一、考試內容與筆記

1. What are the three main characteristics of Object-Oriented Programming?

2. Define in C++ a pointer that points to the element a [5] [10] of the array "int a [10] [20] [30] [40] " named p.

3. Now, assuming that the value of p in Question 2 is z, what would be the values of p + 1 and p - 1?Why?

4. Again, assuming that the value of p in Question 2 is z, what would be the values of (p + 1) - p and p - (p + 1)? Why?

5. Define in C++ a pointer "p" to the array "int a [256] ;" so that p[1] is an alias of a [0], p[2] is an alias of a [1], and so on, all the way up so that p [256] is an alias of a [255].

6. In C++, which four of the operators have to be defined as a member function when overloaded. Why?

7. What would be the output of the following C++ program?

8. What would be the output of the following C++ program?

9. Is the following C++ program correct? If not, show how to correct it.

10. Implement the copy constructor and copy assignment operator for the class in Question 9 so that its instances can only be moved, but not copied.

三、相關文章

三、考試資訊

四、參考資料

留言

張貼留言

［筆記］109-2高等物件導向程式設計期中考