`decltype((variable))` has `const` qualifier inside lambda with capture by value? [duplicate]

#include <iostream> template <class T> void print_type() { std::cout << __PRETTY_FUNCTION__ << std::endl; } int main () { float x, &r = x; std::cout << "capture by value" << std::endl; [=] { print_type<decltype(x)>(); print_type<decltype((x))>(); print_type<decltype(r)>(); print_type<decltype((r))>(); }(); std::cout << "capture by reference" << std::endl; [&] { print_type<decltype(x)>(); print_type<decltype((x))>(); print_type<decltype(r)>(); print_type<decltype((r))>(); }(); } 

The output of clang 10.0 is (GCC 10.0 has a bug):

capture by value void print_type() [T = float] void print_type() [T = const float &] void print_type() [T = float &] void print_type() [T = const float &] capture by reference void print_type() [T = float] void print_type() [T = float &] void print_type() [T = float &] void print_type() [T = float &] 

I wonder why decltype((x)) and decltype((r)) have a const qualifier when they are captured by value. When they are captured by reference, they don't have a const qualifier.

Any idea?

And what's the common use case of decltype((variable)) inside and outside lambda?

Edit#1:

Few flagged this as a duplicate of this.

The post doesn't explain why double parenthesis adds const to decltype's deduction.