如果 constexpr 条件为假，如何避免编译此语句？

如何解决如果 constexpr 条件为假，如何避免编译此语句？

在模板中我可以做到：

struct Type_Which_std_cout_MayAccept {};

template <typename T>
struct AAA
{
    AAA(T t) { if constexpr (false) std::cout << t; }
};

int main()
{
    AAA{ Type_Which_std_cout_MayAccept() };
}

哪个工作正常。这就是为什么我为什么认为如果 constexpr 条件为假，编译器将不会尝试评估该语句。所以我希望这能奏效，但事实并非如此。我显然有错误的理解：

struct NoType {};


template <typename T>
struct MemberOrnothing
{
    MemberOrnothing(T arg) : member(arg) {}
    constexpr bool hasMember() { return true; }
    T member;
};
template <>
struct MemberOrnothing<NoType>
{
    MemberOrnothing(NoType) {}
    constexpr bool hasMember() { return false; }
};

template <typename T_MemberOne,typename T_OptionalSecondMember = NoType>
struct OneOrTwoMemberStruct
{
    OneOrTwoMemberStruct(T_MemberOne firstMember,T_OptionalSecondMember optionalSecond = NoType())
        : firstMember(firstMember),secondMember(optionalSecond)    {}


    T_MemberOne firstMember;
    MemberOrnothing<T_OptionalSecondMember> secondMember;
    T_OptionalSecondMember& getSecondMember() { return secondMember.member; }
    constexpr bool hasSecondMember() { return secondMember.hasMember(); }
    
};

int main()
{

      OneOrTwoMemberStruct objWithSecondMember{ 7,'c' };
      OneOrTwoMemberStruct objWithoutSecondMember{ 7 };

    // Works
    if constexpr (objWithSecondMember.hasSecondMember())
        std::cout << objWithSecondMember.getSecondMember() << '\n';
    
    // Doesn't work,std::cout << doesn't accept a NoType
    if constexpr (false)
        std::cout << objWithoutSecondMember.getSecondMember() << '\n';
} 

在第一个示例中，即使不接受 T，我也可以编写将 T 传递给 cout 的运算符

谁能解释一下两者的区别？