[英]Specialize variadic template member function
Is there a way to get this "fake overloading" argument specialization to work?有没有办法让这种“假重载”参数专业化工作?
#include <iostream>
class Foo
{
public:
template < typename T, typename ... Args >
T* bar (Args&& ... args);
};
template <>
int* Foo::bar()
{
std::cout << "int* bar()\n";
return new int;
}
// error: template-id ‘bar<>’ for ‘double* Foo::bar(double)’ does not match any template declaration
template <>
double* Foo::bar(double d)
{
std::cout << "double* bar(" << d << ")\n";
return new double;
}
int main()
{
Foo f;
f.bar<int>();
f.bar<double>(3.0);
}
One solution would be to actually overload the function, eg一种解决方案是实际使 function 过载,例如
class Foo
{
public:
template < typename T, typename ... Args >
T* bar (Args&& ... args);
template < typename T, typename U >
T* bar (U u);
};
but I would prefer to avoid that if possible.但如果可能的话,我宁愿避免这种情况。 I am looking at dozens of allowed combinations of return type and arguments, so overloading is impractical.
我正在查看返回类型和 arguments 的数十种允许组合,因此重载是不切实际的。
I assume some SFINAE magic should make it possible to only enable very specific combinations of return type and arguments, but I feel that I'm probably missing a simpler option.我假设一些 SFINAE 魔法应该可以只启用返回类型和 arguments 的非常特定的组合,但我觉得我可能缺少一个更简单的选项。
The parameter type of the specialization double
, doesn't match the parameter type of the primary template Args&&
, as the error message said.如错误消息所述,专业化
double
的参数类型与主模板Args&&
的参数类型不匹配。
The primary template is applying forwarding reference, which could instantiate as both lvalue-reference or rvalue-reference.主模板正在应用转发引用,它可以实例化为左值引用或右值引用。 For rvalue-reference (which matches the usage in
main()
) the specialization should be对于右值引用(与
main()
中的用法相匹配),专业化应该是
template <>
double* Foo::bar(double&& d)
{
std::cout << "double* bar(" << d << ")\n";
return new double;
}
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