[英]Partially specializing template class in c++
我有一个模板类:
template <typename T> struct randimpl {
// default is to not compile, have to pass T to something to get assert
// evaluated during second phase expansion, is_void is an arbitrary choice.
static_assert(
std::is_void<T>::value && false, "random() not implemented for type"
);
};
这只是random()
函数的代理,可以为用户定义的类型重载:
// random function, proxies request through randimpl
template <typename T> T random() {
return randimpl<T>::random();
}
我正试图部分专门为类型_point1d<T>
:
// generate random instances _point1d<T>
template <typename T>
struct randimpl<_point1d<T>> {
static _point1d<T> random() {
return _point1d<T>(random<T>());
}
};
据我所知,它是适当的语法,但是当我编译时,我得到:
inc/geometry.h: In static member function ‘static _point1d<T> randimpl<_point1d<T> >::random()’:
inc/geometry.h:287:31: error: expected primary-expression before ‘(’ token
return _point1d<T>(random<T>());
^
inc/geometry.h:287:40: error: expected primary-expression before ‘>’ token
return _point1d<T>(random<T>());
^
inc/geometry.h:287:42: error: expected primary-expression before ‘)’ token
return _point1d<T>(random<T>());
我的语法是否关闭? 这似乎应该有效,但我不明白为什么它不接受它。
问题是static
成员函数的名称random
隐藏了全局函数,它试图调用自身(即递归调用)。 但它不是一个模板,与它的调用方式不符。
您可以添加::
:(范围分辨率)来指定全局random
,例如
// generate random instances _point1d<T>
template <typename T>
struct randimpl<_point1d<T>> {
static _point1d<T> random() {
return _point1d<T>(::random<T>());
// ^^
}
};
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