find out the type of auto

Question

I am playing with generic lambda in C++1y and I often confused by don\'t know what is the type of auto variable/parameter. Is any good way to find it out?

Answer 1

this is what I have ended up with. combined with @Konrad Rudolph's answer and @Joachim Pileborg's comment

std::string demangled(std::string const& sym) {
    std::unique_ptr
    name{abi::__cxa_demangle(sym.c_str(), nullptr, nullptr, nullptr), std::free};
    return {name.get()};
}

template 
void print_type() {
    bool is_lvalue_reference = std::is_lvalue_reference::value;
    bool is_rvalue_reference = std::is_rvalue_reference::value;
    bool is_const = std::is_const::type>::value;

    std::cout << demangled(typeid(T).name());
    if (is_const) {
        std::cout << " const";
    }
    if (is_lvalue_reference) {
        std::cout << " &";
    }
    if (is_rvalue_reference) {
        std::cout << " &&";
    }
    std::cout << std::endl;
};

int main(int argc, char *argv[])
{   
    auto f = [](auto && a, auto b) {
        std::cout << std::endl;
        print_type();
        print_type();
    };

    const int i = 1;
    f(i, i);
    f(1, 1);
    f(std::make_unique(2), std::make_unique(2));
    auto const ptr = std::make_unique();
    f(ptr, nullptr);

}

and output

int const &
int

int &&
int

std::__1::unique_ptr > &&
std::__1::unique_ptr >

std::__1::unique_ptr > const &
std::nullptr_t