模板类中的静态成员初始化 [英] Static member initialization in a template class
问题描述
我需要在模板类内部初始化一个静态布尔,并且尝试像此那样进行操作.我能看到的唯一区别是我对类型参数T有一个约束,但这会导致编译错误,为什么?我该如何解决?
I need to initialize a static bool inside of a template class and I tried to do it like this. The only difference I can see is that I have a constraint on type parameter T but this causes a compilation error, why? How can I solve this?
代码如下:
template<class T, class = typename enable_if<is_any_integral<T>::value>::type>
class fraction {
static bool auto_reduce;
// ...
};
template<class T, class = typename enable_if<is_any_integral<T>::value>::type>
bool fraction<T>::auto_reduce = true;
错误是:
错误:用于声明的嵌套名称说明符'
fraction< T> ::
'未引用类,类模板或类模板的部分专业化布尔分数< T> :: auto_reduce = true;
error: nested name specifier '
fraction<T>::
' for declaration does not refer into a class, class template or class template partial specialization
bool fraction<T>::auto_reduce = true;
推荐答案
也许更简单
template<class T, class V>
bool fraction<T, V>::auto_reduce = true;
撰写时
template<class T, class = typename enable_if<is_any_integral<T>::value>::type>
class fraction
您说 fraction
是具有两个类型模板参数的类; std :: enable_if
(如果该部分可用于为第二个参数分配默认值(并允许启用/不启用SFINAE工作),但是 fraction
是并且仍然是具有两个参数的模板 class
,您必须同时引用这两个参数,并且无需重复启用/不启用/默认部分来初始化第二个参数 auto_reduce
.
you say that fraction
is a class with two type template paramenters; the std::enable_if
if part is useful to assign a default value to the second parameter (and to permit the enable/not enable SFINAE works) but fraction
is and remain a template class
with two parameters, you have to cite both and there is no need to repeat the enable/not enable/default part for second parameter initializing auto_reduce
.
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