c++ - 为什么按值传递给函数和按值传递给另一个构造函数时，构造函数的调用会有所不同？

Question

我正在尝试以下程序:

#include <iostream>
using namespace std;
class Type{
    int i;
    public: 
    Type() {cout << "type constructor "<<endl;}
    Type (const Type &) { cout << "type copy constructor called" << endl;}
};

class MyClass {
    Type variable;
public:
    MyClass(Type a) {   
    cout << "inside MyClass constructor "<<endl;
    variable = a;
    }
};
void fun (Type){
    return;
}

int main (){
    Type t;
    cout <<"t created"<<endl;
    MyClass tmp = MyClass(t);
    cout<<"calling fun"<<endl;
    fun(t);
}

这个的输出是:

type constructor 
t created
type copy constructor called
type constructor 
inside MyClass constructor 
calling fun
type copy constructor called

我想知道为什么当我将它传递给 MyClass 构造函数时调用默认构造函数，为什么当我将它传递给 fun() 时调用复制构造函数？
顺便说一句，当我使用初始化列表时也会发生同样的情况。

Answer 1

I am wondering why default constructor is called when I pass it to MyClass constructor

此处与传参无关。 variable作为成员变量，首先会默认构造。

class MyClass {
    Type variable;  
public:
    MyClass(Type a) {   // variable will be default constructed at first, since it's not initialized via member initializer list
    cout << "inside MyClass constructor "<<endl;
    variable = a;       // then variable is assgined via assignment operator
    }
};

您可以指定成员初始化器列表如何初始化变量，例如

class MyClass {
    Type variable;  
public:
    MyClass(Type a) : variable(a) {   // variable will be direct initialized via copy constructor
    cout << "inside MyClass constructor "<<endl;
    // variable = a;                  // no need for assignment
    }
};

在这种情况下不会调用默认构造函数。