C++构造函数的初始化列表
圣人不入爱河 人气:01.问题
class A { private: int m_a; public: A(int a) { cout << "A(int a)......." << endl; m_a = a; } void print() { cout <<"m_a=" << m_a << endl; } }; class B { private: int m_b; A m_a1; A m_a2; public: B(A& a1,A& a2, int b) { m_b = b; m_a1(a1);//此处调用A的拷贝函数会报错 m_a2(a2);//此处调用A的拷贝函数会报错 } };
错误:
2.解决方法(初始化列表)
将class B构造函数改写为:
public: B(A& a1,A& a2, int b) :m_a1(a1),m_a2(a2)//构造函数的初始化列表 { m_b = b; } };
完整代码如下:
#include <iostream> using namespace std; class A { private: int m_a; public: A(int a) { cout << "A(int a)......." << endl; m_a = a; } void print() { cout <<"m_a=" << m_a << endl; } A(const A& another) { m_a = another.m_a; } ~A() { cout << "~A()......" << endl; } }; class B { private: int m_b; A m_a1; A m_a2; public: B(A& a1,A& a2, int b) :m_a1(a1),m_a2(a2)//构造函数的初始化列表,调用拷贝构造 { cout << "B(A& a1,A& a2, int b)......." << endl; m_b = b; } ~B() { cout << "~B()......." << endl; cout << "m_b=" << m_b << endl; cout << "A m_a1" << endl; m_a1.print(); cout << "A m_a2" << endl; m_a2.print(); } }; int main(int argc, char** argv) { A a1(1), a2(2); B b1(a1, a1, 3); }
运行结果:
3.顺序问题
构造对象成员的顺序跟初始化列表的顺序无关,而是跟成员对象定义的顺序有关。(面试会问)
例子:
class A { private: int m_a; public: A(int a) { cout << "A(int a)......." <<a<< endl; m_a = a; } void print() { cout <<"m_a=" << m_a << endl; } A(const A& another) { m_a = another.m_a; } ~A() { cout << "~A()......"<< endl; } }; class B { private: int m_b; A m_a2; A m_a1; public: B(int a1, int a2, int b) :m_a1(a1), m_a2(a2)//调用有参构造函数 { cout << "B(int a1, int a2, int b)......." << endl; m_b = b; } ~B() { cout << "~B()......." << endl; } }; int main(int argc, char** argv) { B b2(1, 2, 3); }
结果:
跟下面顺序有关:
private: A m_a2; A m_a1;
跟下面顺序无关:
B(int a1, int a2, int b) :m_a1(a1), m_a2(a2)//调用有参构造函数
总结
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