As per the document, which states that[emphasise mine]:

Missing initializers in an initializer list

If an aggregate is initialized but the number of initialization values
is fewer than the number of members, then all remaining members are
initialized with an empty initializer list. In most cases, this will
perform value-initialization on those members.
We have discussed about the quotation here about how to understand it in the right way.

However, the aforementioned statement is not suitable when the class has default initliazers.
Here is the example code below.

#include <iostream>
#include <type_traits>

struct Employee_no_default_member_initializer
{
    int id;
    int age;
    double wage;
};

struct Employee_with_default_member_initializer
{
    int id {1};
    int age {20};
    double wage {5.0};
};

template<typename T>
class is_allowed_type:public std::false_type{};

template<>
class is_allowed_type<Employee_with_default_member_initializer>:public std::true_type{};

template<>
class is_allowed_type<Employee_no_default_member_initializer>:public std::true_type{};

template<typename T, typename std::enable_if<is_allowed_type<T>::value, void>::type* = nullptr>
std::ostream& operator<<(std::ostream&os, const T& employee)
{
    os << "id="<< employee.id << std::endl;
    os << "age="<< employee.age << std::endl;
    os << "wage="<< employee.wage << std::endl;

    return os;
}

template<typename T, typename std::enable_if<is_allowed_type<T>::value,void>::type* = nullptr>
void foo()
{
    T frank = { 3 }; // copy-list initialization using braced list
    T joe { 2, 28};  // list initialization using braced list (preferred)
    std::cout << frank << std::endl;
    std::cout << joe << std::endl;
}

int main()
{
    std::cout << "#1" << std::endl;
    foo<Employee_with_default_member_initializer>();
    std::cout << "#2" << std::endl;
    foo<Employee_no_default_member_initializer>();

    return 0;
}

Here is the output:

#1
id=3
age=20
wage=5

id=2
age=28
wage=5

#2
id=3
age=0
wage=0

id=2
age=28
wage=0