C++ 中的“使用类型”会导致几个错误

Question

In my question type as returntype in c++ I was provided with an answer that gave me such a structure:

template <int N>
struct int_type {
    using type = std::conditional_t<N <= 8, std::uint8_t,
                 std::conditional_t<N <= 16, std::uint16_t,
                 std::conditional_t<N <= 32, std::uint32_t,
                 std::conditional_t<N <= 64, std::uint64_t,
                 std::uintmax_t>>>>;
};

That seemed to do excactly what I need, how ever the practice looks different, since I can't compile it because of the following errors:

...Error: expected nested-name-specifier before 'type'
 using type = std::conditional_t<N <= 8, std::uint8_t,
       ^
...Error: using-declaration for non-member at class scope
...Error: expected ';' before '=' token
 using type = std::conditional_t<N <= 8, std::uint8_t,
            ^
...Error: expected unqualified-id before '=' token

I tried to google it, but none of the posts I found seem to adress this specific problems. Can anyone explain me what is wrong with this code? I'm pretty new to C++

Answer 1

Your code is perfectly legal as long as your compiler supports c++14 stdlib's <type_traits> that define an alias template for std::conditional trait.

However, the error message clearly indicates you've not even enabled c++11 , therefore using is not parsed as an alias at all. To do that, add -std=c++11 or -std=c++14 option to your configuration.

If you can't, then in c++03 you can easily implement your own conditional trait:

template <bool B, typename T, typename F>
struct conditional
{
    typedef T type;
};

template <typename T, typename F>
struct conditional<false, T, F>
{
    typedef F type;
};

template <int N>
struct int_type {
    typedef typename conditional<N <= 8, uint8_t,
                 typename conditional<N <= 16, uint16_t,
                 typename conditional<N <= 32, uint32_t,
                 typename conditional<N <= 64, uint64_t,
                 uintmax_t>::type>::type>::type>::type type;
};

In c++11 you can use std::conditional instead of std::conditional_t , the only difference is that you need to access a nested typedef type on your own, which is a dependent name ( typename keyword needed in front of a nested name specifier ):

#include <cstdint>
#include <type_traits>

template <int N>
struct int_type {
    using type = typename std::conditional<N <= 8, std::uint8_t,
                 typename std::conditional<N <= 16, std::uint16_t,
                 typename std::conditional<N <= 32, std::uint32_t,
                 typename std::conditional<N <= 64, std::uint64_t,
                 std::uintmax_t>::type>::type>::type>::type;
};

In c++11 and c++14 , you need to include <cstdint> instead of <stdint.h> . The former guarantees the names are defined in the std namespace (the latter may define them only in the global namespace).

Answer 2

I'll add an answer here for the sake of completeness.

You need to #include <cstdint> and #include <type_traits> , then compile with C++14 support. Code:

#include <type_traits>
#include <cstdint>
#include <iostream>

template <int N>
struct int_type {
    using type = std::conditional_t<N <= 8, std::uint8_t,
                 std::conditional_t<N <= 16, std::uint16_t,
                 std::conditional_t<N <= 32, std::uint32_t,
                 std::conditional_t<N <= 64, std::uint64_t,
                 std::uintmax_t>>>>;
};

int main()
{
    int_type<10>::type t; // std::uint16_t
    t = 12;
    std::cout << t << std::endl;
}

Live example: http://coliru.stacked-crooked.com/a/9352f3349f48bff4

Answer 3

Huh, when you do a typedef the compiler tells you what's wrong. Those are dependant typenames so you need a "typename" in front of it. so this should work:

#include <iostream>
#include <type_traits>
using namespace std;

template <int N>
struct int_type {
    using mytype = typename std::conditional<N <= 8, std::uint8_t,
                     typename std::conditional<N <= 16, std::uint16_t,
                       typename std::conditional<N <= 32, std::uint32_t,
                         typename std::conditional<N <= 64, std::uint64_t,
                           std::uintmax_t
                         >::type
                       >::type
                     >::type
                   >::type;
};