构造函数的初始化

 C++构造函数初始化是先从初始化列表开始，如果初始化列表没有，就对成员变量调用默认构造函数先进行初始化，再进入构造函数的函数体。

#include <iostream> class Base1 { public: Base1() : i(0) { std::cout << "Base1 default" << std::endl; } Base1(const int& t) : i(t) { std::cout << "construct of t" << std::endl; } Base1(const Base1& t) { i = t.i; std::cout << "copy construct function" << std::endl; } Base1& operator=(const Base1& t) { i = t.i; std::cout << "copy assignment base1" << std::endl; return *this; } Base1& operator=(const int& t) { i = i; std::cout << "copy assignment int" << std::endl; return *this; } private: int i; }; class Base2 { public: Base2() { j = 0; std::cout << "Base2 default construct function" << std::endl; } private: Base1 j; }; int main(void) { Base2 tmp1; std::cout << "tmp1 construct complete" << std::endl; return 0; }

Base2构造函数没有初始化列表，所以先调用Base1的默认构造函数，再进入Base2构造函数的函数体。

#include <iostream> class Base3 { public: Base3() : i(0) { std::cout << "Base3 default" << std::endl; } Base3(const int& t) : i(t) { std::cout << "construct of t" << std::endl; } Base3(const Base3& t) { i = t.i; std::cout << "copy construct function" << std::endl; } Base3& operator=(const Base3& t) { i = t.i; std::cout << "copy assignment Base3" << std::endl; return *this; } Base3& operator=(const int& t) { i = i; std::cout << "copy assignment int" << std::endl; return *this; } private: int i; }; class Base4 { public: Base4() : j(0) { std::cout << "Base4 default construct function" << std::endl; } private: Base3 j; }; int main(void) { Base4 tmp1; std::cout << "tmp1 construct complete" << std::endl; return 0; }

运行结果显示，如果初始化列表进行了初始化，就不会调用默认构造函数再进行初始化，所以通过初始化列表进行初始化提高了代码的效率。