显式复制构造函数编译错误 [英] explicit copy constructor compile error

问题描述

我在C ++中检查操作符重载，遇到了一些我没有想到的并且有一些疑问。

我的拷贝构造函数声明并实现为

 显式Vector（const Vector& v）; 
 
 Vector :: Vector（const Vector&v）：
 _x（v._x），_y（v._y），_z（v._z）{} 
  
 
 
 
然后我重载复合赋值运算符
  Vector Vector :: operator +（const Vector& v）const 
 {
 Vector tmp（* this）; 
 tmp + = v; 
 return tmp; 
} 
 
 Vector Vector :: operator-（const Vector& v）const 
 {
 Vector tmp 
 tmp  -  = v; 
 return tmp; 
} 
  
但是，在 return 语句我得到一个错误说没有匹配的构造函数初始化'Vector'。
 
 
 
我添加到我的构造函数是显式关键字，我删除它和代码编译很好，为什么？
 
 
 
我也在检查C ++ 11的新东西，发生了我可以声明我的构造函数像一个移动构造函数
 显式向量（const Vector& v）; 
  
并且代码编译正常。如果我这样做，我必须有复制和移动构造函数？
 显式Vector（const Vector& v）; 
 explicit Vector（const Vector& v）; 
  
或者只是移动构造函数会正常工作？如果我想坚持C ++ 11，正确的方法是什么？
解决方案
 ，但函数
  Vector Vector :: operator +（const Vector& v）const 
  
 
 
 
和
  Vector Vector :: operator- （const Vector& v）const 
  
必须返回值， $ c> explicit （换句话说，它们不能将 tmp 复制到返回的对象中）。
 
 $ b $我也是从C ++ 11检查新的东西，发生了，我可以声明我的构造函数像一个移动构造函数
 显式Vector（const Vector& ;& v）; ，代码编译正好。如果我这样做，我必须有复制和移动构造函数？
 
 
不知道我明白你的意思。如果你只声明一个显式的移动构造函数（这将阻止编译器生成一个默认的复制构造函数），你会有同样的问题。我不能生成一个可编译代码。 
 
I was checking operator overloading in C++ and came across something I did not expect and have some doubts about it.

My copy constructor is declared and implemented as as
explicit Vector(const Vector& v);

Vector::Vector(const Vector& v) :
_x(v._x), _y(v._y), _z(v._z) {}
then I am overloading the compound assignment operators
Vector Vector::operator+(const Vector& v) const
{
    Vector tmp(*this);
    tmp += v;
    return tmp;
}

Vector Vector::operator-(const Vector& v) const
{
    Vector tmp(*this);
    tmp -= v;
    return tmp;
}
however, in the return statements I got an error saying no matching constructor for initialization of 'Vector'.

Since the only thing I added to my constructor was the explicit keyword, I deleted it and the code compiles just fine, why?

I also was checking new stuff from C++11 and occurred that I can declare my constructor like a moving-constructor
explicit Vector(const Vector&& v);
and the code compiles just fine. If I do that, do I have to have both copy and move constructors?
explicit Vector(const Vector& v);
explicit Vector(const Vector&& v);
or just having the move constructor will work fine? If I want to stick to C++11, what is the correct approach to follow?
解决方案
You defined an explicit copy constructor, but the functions
Vector Vector::operator+(const Vector& v) const
and 
Vector Vector::operator-(const Vector& v) const
must return by value, and cannot anymore, due to explicit (in other words, they cannot copy tmp into the returned object).

  
I also was checking new stuff from C++11 and occurred that I can declare my constructor like a moving-constructor 
  explicit Vector(const Vector&& v); and the code compiles just fine. If I do that, do I have to have both copy and move constructors?
Not sure I understand what you mean here. You will have the same issue if you only declare an explicit move constructor (which will prevent the compiler from generating a default copy constructor). I am not able to produce a "compilable" code. 

                        
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