澄清了当2个const int相乘时的decltype输出

int main()
{
 const int a = 1;
 const int b = 2;
 typedef decltype(a*b) multiply_type;
 cout << typeid(multiply_type).name() << endl;
 return 0;
}

程序的返回值是multiply_type为int。我很惊讶。我期望类型推导产生const int，由于表达式产生一个pr值，因此结果类型将是const int。

PS:对于auto，返回值将是int，因为它去掉了const限定符。

有没有人知道为什么multiply_type是int而不是const int与decltype ?

编辑:添加了一个额外的例子，也与cv-qualifier有关。

#include<iostream>
#include<typeinfo>

using namespace std;
struct Details
{
    int m_age;
};
int main()
{
 const Details* detail = new Details();
 typedef decltype((detail->m_age)) age_type;
 cout << typeid(age_type).name() << endl;
 int a = 1;
 age_type age = a;
 age = 10; // This is not possible. Read only. 
 cout << typeid(age).name() << endl; // This returns the type as int though. Then why is 20 not possble ?
 return 0;
}

编辑2:检查我们的链接。http://thbecker.net/articles/auto_and_decltype/section_07.html'

int x;
const int& crx = x;
/ The type of (cx) is const int. Since (cx) is an lvalue,
// decltype adds a reference to that: cx_with_parens_type
// is const int&.
typedef decltype((cx)) cx_with_parens_type;` 

using T = const int;
static_assert(is_same<int, decltype(0)>(), "Failed");