更新时间:2022-04-10 09:49:54
是的,显式专门化函数而不完全专门化所有外部模板是不可能的函数 - 它周围不能有任何可变部分,仍然由模板参数化)
Yes, explicitly specializing a function without fully specializing all outer template is not possible (an explicit function specialization is a real function - there can't be any "variable parts" around it that are still parameterized by a template)
一个简单的方法是使用type2type模板和重载:
A simple way is to use a type2type template together with overloading:
template<typename T> struct t2t { typedef T type; };
void Func( Parm1 arg1, Parm2, arg2 ) { Call<Parm3>(arg1, arg2, t2t<Parm3>()); }
template< class Type, class V > void Call( Parm1 arg1, Parm2 arg2, t2t<V>) { }
template< class Type > void Call( Parm1 arg1, Parm2 arg2, t2t<void>) { }
第二个调用
重载,如果你用 t2t< void>
调用它,否则,因为第一个
Now, it will call the second Call
overload if you call it with t2t<void>
, and the first otherwise, because the first one is less special.
使用 enable_if
:
void Func( Parm1 arg1, Parm2, arg2 ) { Call<Parm3>(arg1, arg2); }
template< class Type > typename disable_if< is_same<Type, void> >::type
Call( Parm1 arg1, Parm2 arg2) { }
template< class Type > typename enable_if< is_same<Type, void> >::type
Call( Parm1 arg1, Parm2 arg2) { }
如果 Type
是void,则取第二个值,如果 Type
又是其他值,则取第一个值。但使用不同的技术。这个被称为 SFINAE
。另一种方法,但它又添加了一个参数,这是为了演示SFINAE的工作原理:
Now, the second one is taken if Type
is void, and the first one is taken if Type
is something else again. But using a different technique. This one is called SFINAE
. An alternative way, but which again adds one parameter is this - to demonstrate the way SFINAE works:
void Func( Parm1 arg1, Parm2, arg2 ) { Call<Parm3>(arg1, arg2); }
template< class Type >
void Call( Parm1 arg1, Parm2 arg2, char(*)[!is_same<Type, void>::value] = 0) { }
template< class Type >
void Call( Parm1 arg1, Parm2 arg2, char(*)[ is_same<Type, void>::value] = 0) { }
如果模板参数的替换产生无效的类型或构造,则会发生SFINAE
。下面,我们尝试创建一个指向0或1的数组的指针。大小为0的数组无效,并且将导致SFINAE失败 - 如果相应的模板专用化是函数,则不会将其视为调用候选。
SFINAE
happens if the substitution of a template parameter yields to an invalid type or construct. Below, we try to create a pointer to an array of size 0 or 1 respectively. An array of size 0 is not valid, and will cause an SFINAE failure - the corresponding template specialization will not be considered as a call-candidate if it is a function.
在上面的 enable_if
中,它的工作原理不同。如果 enable_if
给出从 false_type
派生的东西,那么它使它的 :: type
typedef不存在。
is_same
从 false_type
中导出类型不同的情况。然后,我们将尝试访问一个不存在的名称 - 这是一个无效的结构,因此也将是一个SFINAE失败。
In the enable_if
case above, it works different. If enable_if
is given something derived from false_type
, then it makes its ::type
typedef not existent. is_same
derives itself from false_type
in the case types are not the same. We would then try to access a not existent name - which is an invalid construct and would therefor be an SFINAE failure too.