我有一种情况,似乎没有调用构造函数：
- #include <iostream>
- using namespace std;
- int main ()
- {
- class yoyo
- {
- public:
- int i;
- yoyo()
- {
- i = 0;
- cout << "defaultly initialized to 0" << endl;
- }
- yoyo (int j) : i(j)
- {
- cout << "initialized to " << j << endl;
- }
- };
- int i;
- yoyo a;
- cout << "Hello1,i: " << a.i << endl;
- yoyo b(5);
- cout << "Hello2,i: " << b.i << endl;
- yoyo c = b; /* 1 */
- cout << "Hello3,i: " << c.i << endl;
- return 0;
- }
输出是：
- defaultly initialized to 0
- Hello1,i: 0
- initialized to 5
- Hello2,i: 5
- Hello3,i: 5
(注意：Hello2和Hello3之间没有任何内容)
如果我将程序更改为如下所示：
- #include <iostream>
- using namespace std;
- int main ()
- {
- class yoyo
- {
- public:
- int i;
- yoyo()
- {
- i = 0;
- cout << "defaultly initialized to 0" << endl;
- }
- yoyo (int j) : i(j)
- {
- cout << "initialized to " << j << endl;
- }
- };
- int i;
- yoyo a;
- cout << "Hello1,i: " << b.i << endl;
- yoyo c; c = b; /* 1 */
- cout << "Hello3,i: " << c.i << endl;
- return 0;
- }
(唯一的区别在于他标有/ * 1 * /的行)
现在的输出是：
- defaultly initialized to 0
- Hello1,i: 5
- defaultly initialized to 0
- Hello3,i: 5
解决方法
如果是
- yoyo c = b;
它被称为copy constructor.
而在这种情况下
- yoyo c; c = b;
这是被称为copy assignment的操作符.
如果您不提供其中任何一个,编译器将为您生成默认版本.
如果要创建自己的复制构造函数,它可能如下所示：
- yoyo(const yoyo& other)
- : i(other.i)
- { std::cout << "copy constructor initialized\n"; }
复制赋值运算符如下所示：
- yoyo& operator=(const yoyo& other)
- {
- i = other.i;
- return *this;
- }
当然,它们都在类定义中定义.
