c++ - 声明成员对象而不调用其默认构造函数

Question

我有两个类:Generator 和 Motor。这是 Generator 的精简版本:

class Generator {
private:
    Motor motor;    // this will be initialized later
                    // However, this line causes the constructor to run.
public: 
    Generator(string);
}

Generator::Generator(string filename) {
    motor = Motor(filename);     // initialisation here
                                 
}

这是 Motor 类定义:

class Motor {
public:
    Motor();
    Motor(string filename);
}

Motor::Motor() {
    cout << "invalid empty constructor called" << endl;
}
Motor::Motor(string filename) {
    cout << "valid constructor called" << endl;
}

这是我的 main() 函数:

int main(int argc, char* argv[]) {
    Generator generator = Generator(argv[1]);
    ....
    ....
}

输出为

invalid empty constructor called

valid constructor called

如何定义类 Generator 以获得 Motor 的实例，而不调用 Motor 的空构造函数，直到稍后？

我必须包含空构造函数，因为如果没有它，g++ 拒绝编译。

Answer 1

您需要使用 intialization list在 Generator 构造函数中进行构造:

Generator::Generator(string filename) : motor(filename) // initialization here
{
    // nothing needs to do be done here
}

您的原始代码实际上在做什么:

Generator::Generator(string filename) /* : motor() */ // implicit empty initialization here
{
    motor = Motor(filename) // create a temp instance of Motor
    //    ^------------------- copy it into motor using the default operator=()
                            // destruct the temp instance of Motor
}