是否有理由使用std :: conjunction/std :: disjunction而不是在“&///||"上的折叠表达式? [英] Is there a reason to use std::conjunction/std::disjunction instead of a fold expression over "&&"/"||"?
问题描述
在某些特殊情况下,您不能正确地使用 std :: conjunction
/ std :: disjunction
而不使用更多的基本"(即语言功能而不是库功能)在&&
/ ||
上折叠表达式?
Is there any specific cases you cannot correctly do with std::conjunction
/std::disjunction
and not using the more "fundamental" (i.e. language feature instead of library feature) fold expression over &&
/||
?
示例:
// func is enabled if all Ts... have the same type
template<typename T, typename... Ts>
std::enable_if_t<std::conjunction_v<std::is_same<T, Ts>...> >
func(T, Ts...) {
// TODO something to show
}
vs
// func is enabled if all Ts... have the same type
template<typename T, typename... Ts>
std::enable_if_t<(std::is_same<T, Ts> &&...)>
func(T, Ts...) {
// TODO something to show
}
使用fold表达式的版本更简短,通常更易读(尽管对此可能有不同的看法).因此,我不明白为什么将它与折叠表达式一起添加到库中.
The version using a fold expression is more brief and generally more readable (although opinions might differ on that). So I don't see why it was added to the library together with fold expressions.
推荐答案
std :: conjunction
短路了 :: value
实例化,而fold表达式没有.这意味着,鉴于:
std::conjunction
short-circuits ::value
instantiation, while the fold expression doesn't. This means that, given:
template <typename T>
struct valid_except_void : std::false_type { };
template <>
struct valid_except_void<void> { };
以下内容将编译:
template <typename... Ts>
constexpr auto test = std::conjunction_v<valid_except_void<Ts>...>;
constexpr auto inst = test<int, void>;
但是以下内容不会:
template <typename... Ts>
constexpr auto test = (valid_except_void<Ts>::value && ...);
constexpr auto inst = test<int, void>;
来自 cppreference :
连接是短路的:如果存在带有
bool(Bi :: value)== false
的模板类型参数Bi
,则实例化conjunction<B1,...,BN> :: value
不需要实例化j>的
.Bj :: value
.我
Conjunction is short-circuiting: if there is a template type argument
Bi
withbool(Bi::value) == false
, then instantiatingconjunction<B1, ..., BN>::value
does not require the instantiation ofBj::value
forj > i
.
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