C++构造函数的初始化列表详解

1.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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