1. Two special constructors
- no-argument constructor has no arguments constructor to explain: no-argument constructor appears to be very simple, but it is especially in it must exist . Because we want to use a class, you must create the object. Create objects necessarily involves calling the constructor. NOTE: When a class is not defined when the constructor, the compiler provided by default a constructor with no arguments, and the function body is empty. Must pay special attention when a class has no constructor, the compiler will provide, or will not provide. - copy constructor constructor parameter const class_name & a (Analyzing a constructor is not based on the copy constructor: . 1, const 2, refer to the current class of objects ) as long as that of such parameters appear in a constructor function, it must It is a copy constructor up.
When a class is not defined copy constructor, the compiler provides a copy constructor, simply copying a value of the member variables default.
1 #include <stdio.h>
2
3 class Test
4 {
5 private:
6 int i;
7 int j;
8 public:
9 int getI()
10 {
11 return i;
12 }
13 int getJ()
14 {
15 return j;
16 }
17 /*Test(const Test& t)
18 {
19 i = t.i;
20 j = t.j;
21 }
22 Test()
23 {
24 }*/
25 };
26
27 int main()
28 {
29 Test t1;
30 Test t2 = t1;
31
32 printf("t1.i = %d, t1.j = %d\n", t1.getI(), t1.getJ());
33 printf("t2.i = %d, t2.j = %d\n", t2.getI(), t2.getJ());
34
35 return 0;
36 }
In the above example, once defined
Test(const Test& t) { i = t.i; j = t.j; }
那么Test t1;就会出错。为什么？
因为找不到一个Test()的构造函数。
问题：不是编译器会默认提供一个无参构造函数吗？
一定要注意，编译器提供默认构造函数的前提，就是类中没有构造函数，一旦拥有，编译器就不再提供。
同样，如果定义了
Test()
{
}
Test t2 = t1，也会出错。原因和上面一样。
呈现一个经典面试题，下面的这个类，它里面有什么东西？
class T
{
};
它里面至少有一个无参构造函数，是编译器提供的。
2. 拷贝构造函数的意义
-兼容C语言的初始化方式
例如：int i =2; int j = i;
Test t1; Test t2 = t1;
-初始化行为能够符合预期的逻辑
-浅拷贝
拷贝后对象的物理状态相同
-深拷贝
拷贝后对象的逻辑状态相同
编译器提供的拷贝构造函数只进行浅拷贝。
1 #include <stdio.h>
2
3 class Test
4 {
5 private:
6 int i;
7 int j;
8 int* p;
9 public:
10 int getI()
11 {
12 return i;
13 }
14 int getJ()
15 {
16 return j;
17 }
18 int* getP()
19 {
20 return p;
21 }30 Test(int v)
31 {
32 i = 1;
33 j = 2;
34 p = new int;
35
36 *p = v;
37 }42 };
43
44 int main()
45 {
46 Test t1(3);
47 Test t2(t1);
48
49 printf("t1.i = %d, t1.j = %d, *t1.p = %p\n", t1.getI(), t1.getJ(), t1.getP());
50 printf("t2.i = %d, t2.j = %d, *t2.p = %p\n", t2.getI(), t2.getJ(), t2.getP());54
55 return 0;
56 }
上面的这个程序，编译会顺利通过。并且你会看到一个现象，t1对象的p和t2对象的p指向了同一段堆空间。你是不是想，这是肯定的啊。因为我就是用t1初始化t2，这正是我想要的结果。但是根据经验，p是在堆上生成的，当我们不用的时候，需要将其释放掉。如下面的代码所示：
1 #include <stdio.h>
2
3 class Test
4 {
5 private:
6 int i;
7 int j;
8 int* p;
9 public:
10 int getI()
11 {
12 return i;
13 }
14 int getJ()
15 {
16 return j;
17 }
18 int* getP()
19 {
20 return p;
21 }30 Test(int v)
31 {
32 i = 1;
33 j = 2;
34 p = new int;
35
36 *p = v;
37 }
38 void free()
39 {
40 delete p;
41 }
42 };
43
44 int main()
45 {
46 Test t1(3);
47 Test t2(t1);
48
49 printf("t1.i = %d, t1.j = %d, *t1.p = %p\n", t1.getI(), t1.getJ(), t1.getP());
50 printf("t2.i = %d, t2.j = %d, *t2.p = %p\n", t2.getI(), t2.getJ(), t2.getP());
51
52 t1.free();
53 t2.free();
54
55 return 0;
56 }
编译就会出错，因为我们释放了两次相同堆空间中的内存。t1中的p和t2中的p指向了同一段堆空间。
如何解决这个问题？
应该手工来提供一个拷贝构造函数
1 #include <stdio.h>
2
3 class Test
4 {
5 private:
6 int i;
7 int j;
8 int* p;
9 public:
10 int getI()
11 {
12 return i;
13 }
14 int getJ()
15 {
16 return j;
17 }
18 int* getP()
19 {
20 return p;
21 }
22 Test(const Test& t)
23 {
24 i = t.i; //首先是值的复制
25 j = t.j;
26 p = new int;
27
28 *p = *t.p; //p的值就不能直接复制了，需要重新到堆空间中申请。申请完之后，将t1对象中p空间中的值拿出来，放到新申请的堆空间中去。
29 }
30 Test(int v)
31 {
32 i = 1;
33 j = 2;
34 p = new int;
35
36 *p = v;
37 }
38 void free()
39 {
40 delete p;
41 }
42 };
43
44 int main()
45 {
46 Test t1(3);
47 Test t2(t1);
48
49 printf("t1.i = %d, t1.j = %d, *t1.p = %p\n", t1.getI(), t1.getJ(), t1.getP());
50 printf("t2.i = %d, t2.j = %d, *t2.p = %p\n", t2.getI(), t2.getJ(), t2.getP());
51
52 t1.free();
53 t2.free();
54
55 return 0;
56 }