模板上的重载算术运算符导致未解决的外部错误 [英] Overloaded arithmetic operators on a template causing an unresolved external error
问题描述
假设我在 Vector4.h 中有这个类:
:
#pragma once
template<typename T>
class Vector4
{
public:
T X;
T Y;
T Z;
T W;
Vector4();
Vector4(T X, T Y, T Z, T W);
~Vector4();
friend Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r);
};
#include "Vector4.inl"
.inl :
and in Vector4.inl:
template<typename T>
Vector4<T>::Vector4()
{
X = 0;
Y = 0;
Z = 0;
W = 0;
}
template<typename T>
Vector4<T>::Vector4(T X, T Y, T Z, T W)
{
this->X = X;
this->Y = Y;
this->Z = Z;
this->W = W;
}
template<typename T>
Vector4<T>::~Vector4()
{
}
template<typename T>
Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r)
{
return(Vector4<T>(l.X * r.X, l.Y * r.Y, l.Z * r.Z, l.W * r.W));
}
当我使用它在这样的地方:
And when I use it somewhere like this:
Vector4<float> a, b;
a = a * b;
它给我一个 LNK2019未解决的外部符号
推荐答案
如注释中所述,你的朋友函数声明
As stated in the comments, your friend function declaration
friend Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r);
在全局命名空间中声明一个非模板函数。当您实例化 Vector4< int>
,函数
declares a non-template function in the global namespace. When you instantiate e.g. Vector4<int>
, the function
Vector4<int> operator*(const Vector4<int>& l, const Vector4<int>& r)
<被声明。请注意,它不是 功能模板。 (另见[temp.friend])
is declared. Note that it's not a function template. (Also see [temp.friend])
然后您的 Vector4.inl
声明并定义函数模板
Your Vector4.inl
then declares and defines a function template
template<typename T>
Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r)
<即一个过载的前一个函数。在表达式 a * b
中,重载解析在模板版本上选择非模板 operator *
.match.best] / 1)。这会导致链接器错误,因为尚未定义非模板函数。
i.e. an overload of the former function. In the expression a * b
, overload resolution chooses the non-template operator*
over the template version (see [over.match.best]/1). This results in a linker error, as the non-template function hasn't been defined.
愚弄自己,简短说明:
template<typename T>
Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r);
由于此运算符是一个自由函数(非成员函数),所以这两行声明一个函数模板,很像
As this operator is a free function (a non-member function), these two lines declare a function template, much like
template<typename T>
Vector4<T> wup();
另一方面,
template<typename T>
Vector4<T> Vector4<T>::operator*(const Vector4<T>& r)
{ /* ... */ }
定义类模板( Vector4
)的成员函数(非模板)。
defines a member function (non-template) of a class template (Vector4
).
一个解决方案是使用前缀声明,并仅与特定的专业化:
One solution is to use forward-declarations and befriend only a specific specialization:
template<typename T>
class Vector4;
template<typename T>
Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r);
template<typename T>
class Vector4
{
public:
T X;
T Y;
T Z;
T W;
Vector4();
Vector4(T X, T Y, T Z, T W);
~Vector4();
// compiler knows of some function template `operator*`,
// can name an specialization:
// ~~~~~~~~~~~~~~~~~~~~~~~~vvv
friend Vector4<T> operator*<T>(const Vector4<T>& l, const Vector4<T>& r);
};
template<typename T>
Vector4<T>::Vector4()
{
X = 0;
Y = 0;
Z = 0;
W = 0;
}
template<typename T>
Vector4<T>::Vector4(T X, T Y, T Z, T W)
{
this->X = X;
this->Y = Y;
this->Z = Z;
this->W = W;
}
template<typename T>
Vector4<T>::~Vector4()
{}
template<typename T>
Vector4<T> operator*(const Vector4<T>& l, const Vector4<T>& r)
{
return(Vector4<T>(l.X * r.X, l.Y * r.Y, l.Z * r.Z, l.W * r.W));
}
int main()
{
Vector4<float> a, b;
a = a * b;
}
另一个解决方案是让整个
模板而不是单个专门化:
Another solution would be to friend the whole operator*
template instead of a single specialization:
template<typename U>
friend Vector4<U> operator*(Vector4<U> const&, Vector4<U> const&);
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