C ++“没有用于初始化的匹配构造函数”编译器错误
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C++ "No matching constructor for initialization of" compiler error
提问by Xavier
I have a class which I try to initialize but get the error "No matching constructor for initialization of 'TextureCoordinates'";
我有一个类，我尝试初始化但收到错误“没有匹配的构造函数用于初始化 'TextureCoordinates'”；
Class which I'm trying to initialize:
我试图初始化的类：
class TextureCoordinates
{
public:
TextureCoordinates(){};
TextureCoordinates(Point2D& origin, Dimensions2D& dim);
Point2D getOrigin() const {return origin;};
Dimensions2D getDim() const {return dim;};
private:
Point2D origin;
Dimensions2D dim;
};
Line with compiler error:
带有编译器错误的行：
TextureCoordinates result(point, Dimensions2D(width, height));
Definition of constructor:
构造函数定义：
TextureCoordinates::TextureCoordinates(Point2D& origin, Dimensions2D& dim):
origin(origin), dim(dim) {}
Any ideas what I'm doing wrong?
任何想法我做错了什么？
回答by celtschk
Your constructor takes the arguments by non-const reference, but you pass a temporary object (Dimensions2D(width, height)) to it. Temporaries, even non-const ones, do not bind to non-const references.
您的构造函数通过非常量引用获取参数，但您将临时对象 ( Dimensions2D(width, height))传递给它。临时对象，即使是非常量的，也不会绑定到非常量引用。
Solution, make your constructor take const references (it shouldn't modify the passed objects anyway):
解决方案，让你的构造函数接受 const 引用（它不应该修改传递的对象）：
TextureCoordinates(Point2D const& origin, Dimensions2D const& dim);
回答by jpalecek
TextureCoordinates result(point, Dimensions2D(width, height))
The second parameter is an rvalue that cannot be bound to lvalue reference the constructor expects:
第二个参数是一个不能绑定到构造函数期望的左值引用的右值：
TextureCoordinates(Point2D& origin, Dimensions2D& dim);
You can fix it by changing the signature of the constructor to
您可以通过将构造函数的签名更改为
TextureCoordinates(Point2D& origin, const Dimensions2D& dim);
TextureCoordinates(Point2D& origin, Dimensions2D&& dim); // alternative for c++11
(if you can) or making the parameter a variable
（如果可以）或使参数成为变量
Dimension2D dim=Dimensions2D(width, height);
TextureCoordinates result(point, dim)
回答by neagoegab
Declare Dimensions2d outside.
在外面声明 Dimensions2d。
Dimension2d d(width, height);
TextureCoordinates result(point, d);
回答by bhuwansahni
Temporary variables cannot be passed as a reference in C++ because then you can change the value of a temporary object in the function that is soon going to disappear!! No such problem exists for reference to constants..... So your function definition should be like
临时变量不能作为 C++ 中的引用传递，因为这样您就可以在即将消失的函数中更改临时对象的值！！引用常量不存在这样的问题......所以你的函数定义应该像
TextureCoordinates(Point2D const& origin, Dimensions2D const& dim);
回答by Hicham
did you write the implementation of :
您是否编写了以下实现：
TextureCoordinates(Point2D& origin, Dimensions2D& dim);
