{ typedef std::common_type_t<char,short,int> A; // int typedef std::common_type_t<float,double> B; // double typedef std::common_type_t<Derived,Base> C; // Base typedef std::common_type_t<Derived*,Base*> D; // Base* typedef std::common_type_t<const int,volatile int> E;...
否则,若std::decay<decltype(false?std::declval<constT1&>():std::declval<constT2&>())>::type为合法类型,则成员type代表该类型; (C++20 起) 否则,无成员type。 若sizeof...(T)大于二(即T...由类型T1, T2, R...组成),则若std::common_type<T1, T2>::type存在,则成员type指代std::common...
但是，返回类型是Matrix<T>，您返回Matrix<C>。所以我建议你这样写(别忘了相应地调整声明)：...
#include <type_traits> template <typename T, typename U> typename std::common_type<T, U>::type max(T a, U b) { return b < a ? a : b; } 上面的函数, 使用下面的代码输出 Copy Highlighter-hljs/// 代码1 std::cout << tt::max(1.3, 2) << "\n\n"; /// 代码2 std::co...
T > struct common_type; (since C++11) Determines the common type among all types T..., that is the type all T... can be implicitly converted to. If such a type exists (as determined according to the rules below), the member type names that type. Otherwise, there is no member ...
给定引用类型是, common_reference 试图寻找所有提供的引用类型都能绑定到的引用类型,但若找不到这种引用类型,则可能返回非引用类型。 若sizeof...(T) 为零,则无成员 type。 若sizeof...(T) 为一(即 T... 仅含一个类型 T0),则成员 type 指名与 T0 相同的类型。 若sizeof...(T) 为二(即 T....
std::common_type(std::chrono::duration) std::common_type(std::chrono::time_point) std::conditional std::conjunction std::const_mem_fun1_ref_t std::const_mem_fun1_t std::const_mem_fun_ref_t std::const_mem_fun_t std::const_pointer_cast std::cref std::ctime std::current_exception...
It also doesn’t help to give arg1 and arg2 a common type T: 即使arg1和arg2的具有相同的模板参数类型T,也无济于事: template<typename T>boolisless(T arg1, T arg2) {returnarg1 <arg2; } because then the compiler gets conflicting types when trying to deduce T for arg1 and arg2. ...
constexprstd::common_type_t<M, N>gcd(M m, N n); (library fundamentals TS v2) Computes the greatest common divisor of the integersmandn. Parameters m, n-integer values Return value If bothmandnare zero, returns zero. Otherwise, returns the greatest common divisor of|m|and|n|. ...
ofType<T>() 仅匹配 T 类型的值 - argThat(predicate: (T) -> Bool) 仅匹配由 predicate 筛选出的 T 类型的值 - none() 匹配选项类型的 None 值 - 如果使用单个标识符作为参数匹配器,则优先选择结构相等的参数。 如果方法有默认参数,并且没有显式指定该参数,则使用 any() 匹配器。 示例: let p =...