[英]How to specialize templated operator overloads?
I'm trying to overload comparison operators as non-members for a particular templated class sub
, 1) between instances of sub
, and 2) between sub
and a specific variable, which returns an instance of either comparer_sub
for the first case and comparer_el
in the alternative, which perform comparisons on sub
along with some other useful members:我正在尝试将比较运算符重载为特定模板化类
sub
非成员,1) 在sub
实例之间,以及 2) 在sub
和特定变量之间,它返回一个comparer_sub
的实例作为第一种情况和comparer_el
in替代方法,它与其他一些有用的成员一起对sub
进行比较:
template <typename T1, typename T2>
class sub_base {
public:
sub_base() {};
};
template <typename T>
class mat {
class sub : public sub_base<T,mat<T>> {
public:
sub(): sub_base<T,mat<T>>() {};
};
public:
int mb;
sub getSub() {return sub();};
};
template <typename T1,typename T2>
class comparer_base {
public:
comparer_base() {};
void base_method() {};
};
template <typename lT1,typename lT2,typename rT1,typename rT2>
class comparer_sub : public comparer_base<sub_base<lT1,lT2>,sub_base<rT1,rT2>> {
using comparer_base<sub_base<lT1,lT2>,sub_base<rT1,rT2>>::base_method;
public:
comparer_sub() : comparer_base<sub_base<lT1,lT2>,sub_base<rT1,rT2>>() {};
};
template <typename lT1,typename lT2,typename rT>
class comparer_el : public comparer_base<sub_base<lT1,lT2>,rT> {
using comparer_base<sub_base<lT1,lT2>,rT>::base_method;
public:
comparer_el() : comparer_base<sub_base<lT1,lT2>,rT>() {};
};
template <typename lT1,typename lT2,typename rT1,typename rT2>
comparer_sub<lT1,lT2,rT1,rT2> operator== (const sub_base<lT1,lT2>& lhs, const sub_base<rT1,rT2>& rhs){
printf("comparer_sub\n");
return comparer_sub<lT1,lT2,rT1,rT2>();
};
template <typename lT1,typename lT2,typename rT>
comparer_el<lT1,lT2,rT> operator== (const sub_base<lT1,lT2>& lhs, const rT& rhs){
printf("comparer_el\n");
return comparer_el<lT1,lT2,rT>();
};
However, any type of comparison I try to perform the second overload is called, what I assume is due to const rt&
being too generic and any instance of sub
fits that argument.但是,我尝试执行第二个重载的任何类型的比较都会被调用,我认为这是由于
const rt&
过于通用,并且sub
任何实例都适合该参数。
int main(int argc, char const *argv[]) {
mat<int> A;
mat<float> B;
float C = 0.6;
A.getSub() == B.getSub(); // comparer_el
A.getSub() == C; // comparer_el
return 0;
}
How can I force the first overload to be prioritized over the second overload between instances of sub
??如何在
sub
实例之间强制第一个过载优先于第二个过载?
The problem is, comparer_el
with rT=sub
is a better choise than the substitution+cast from sub
to sub_base<rT1,rT2>
.问题是,
comparer_el
与rT=sub
比替代+铸造从更好的choise sub
到sub_base<rT1,rT2>
Deduction does not implicitly cast (except putting additional c/v qualifiers)... For a much better description, read this answer .演绎不会隐式转换(除了放置额外的 c/v 限定符)......要获得更好的描述,请阅读此答案。
So, one solution is not to use return type sub
within getSub()
but the base type sub_base<rT1,rT2>
.因此,一种解决方案不是在
getSub()
使用返回类型sub
,而是使用基本类型sub_base<rT1,rT2>
。 see here看这里
important part:重要部分:
template <typename T>
class mat {
class sub : public sub_base<T,mat<T>> {
public:
sub(): sub_base<T,mat<T>>() {};
};
public:
int mb;
sub_base<T,mat<T>> getSub() {return sub();};
// ^^^^^^^^^^^^^^^^^^ return type changed from sub to sub_base<T,mat<T>>
};
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