Say I have
constexpr const std::uint8_t major = 1;
constexpr const std::uint8_t minor = 10;
constexpr const std::uint8_t bugfix = 0;
and I want
constexpr const char* version_string(){ ... }
to return the equivalent of "1.10.0"
in this example, how would I do it?
I assume I'll need both of these, in constexpr
:
The problem is purely academic, and I see little to no use to actually have it constexpr
other than "it's possible". I just can't see how this would pan out. I'm willing to accept C++1y solutions that work on GCC 4.9 and Clang 3.4/3.5.
I believe I have found nearly what I seek on some Japanese blogs:
constexpr itoa
constexpr strcat
I will see what I can do with these, and perhaps answer this self-declared interesting question myself when I'm satisfied with the result.
Here's a little C++1y solution --- I think I LOVE C++1y.
#include <utility>
template<int N>
struct c_string
{
int length;
char str[N+1];
constexpr explicit c_string(int p_length)
: length(p_length), str{}
{}
};
template<int M>
constexpr auto make_c_string(char const (&str)[M])
{
c_string<M-1> ret{M-1};
for(int i = 0; i < M; ++i)
{
ret.str[i] = str[i];
}
return ret;
}
template<int N, int M>
constexpr auto join(c_string<N> const& x, c_string<M> const& y)
{
c_string<N+M> ret{x.length + y.length};
for(int i = 0; i < x.length; ++i)
{
ret.str[i] = x.str[i];
}
for(int i = 0; i < y.length; ++i)
{
ret.str[i+x.length] = y.str[i];
}
ret.str[N+M] = '\0';
return ret;
}
template<int N, int M>
constexpr auto operator+(c_string<N> const& x, c_string<M> const& y)
{
return join(x, y);
}
template<class T>
constexpr void c_swap(T& x, T& y)
{
T tmp( std::move(x) );
x = std::move(y);
y = std::move(tmp);
}
// from http://en.cppreference.com/w/cpp/algorithm/reverse
template<class I>
constexpr void reverse(I beg, I end)
{
while(beg != end && beg != --end)
{
c_swap(*beg, *end);
++beg;
}
}
Now the constexpr itoa
:
#include <limits>
template<class T>
constexpr auto c_abs(T x)
{
return x < T{0} ? -x : x;
}
template<class T>
constexpr auto ntoa(T n)
{
c_string< std::numeric_limits<T>::digits10 + 1 > ret{0};
int pos = 0;
T cn = n;
do
{
ret.str[pos] = '0' + c_abs(cn % 10);
++pos;
cn /= 10;
}while(cn != T{0});
if(n < T{0})
{
ret.str[pos] = '-';
++pos;
}
ret.str[pos] = '\0';
ret.length = pos;
reverse(ret.str, ret.str+ret.length);
return ret;
}
We can then simplify the usage:
#include <type_traits>
// not supported by the libstdc++ at coliru
//template<class T, class = std::enable_if_t< std::is_arithmetic<T>{} >>
template<class T, class = typename std::enable_if<std::is_arithmetic<T>{}>::type>
constexpr auto to_c_string(T p)
{
return ntoa(p);
}
template<int N>
constexpr auto to_c_string(char const (&str)[N])
{
return make_c_string(str);
}
template<class T, class U, class... TT>
constexpr auto to_c_string(T&& p0, U&& p1, TT&&... params)
{
return to_c_string(std::forward<T>(p0))
+ to_c_string(std::forward<U>(p1), std::forward<TT>(params)...);
}
And a usage example:
#include <iostream>
int main()
{
constexpr auto res = to_c_string(42," is the solution, or is it ",-21,"?");
std::cout << res.str;
}
Live example @ coliru's clang++3.4
Here is a C++11 solution. It uses class templates with char...
parameter pack to simulate strings:
#include <iostream>
#include <type_traits>
template <char... symbols>
struct String
{
static constexpr char value[] = {symbols...};
};
template <char... symbols>
constexpr char String<symbols...>::value[];
template <typename, typename>
struct Concat;
template <char... symbols1, char... symbols2>
struct Concat<String<symbols1...>, String<symbols2...>>
{
using type = String<symbols1..., symbols2...>;
};
template <typename...>
struct Concatenate;
template <typename S, typename... Strings>
struct Concatenate<S, Strings...>
{
using type = typename Concat<S, typename Concatenate<Strings...>::type>::type;
};
template <>
struct Concatenate<>
{
using type = String<>;
};
template <std::size_t N>
struct NumberToString
{
using type = typename Concat
<
typename std::conditional<(N >= 10), typename NumberToString<N / 10>::type, String<>>::type,
String<'0' + N % 10>
>::type;
};
template <>
struct NumberToString<0>
{
using type = String<'0'>;
};
constexpr const std::uint8_t major = 1;
constexpr const std::uint8_t minor = 10;
constexpr const std::uint8_t bugfix = 0;
using VersionString = Concatenate
<
NumberToString<major>::type,
String<'.'>,
NumberToString<minor>::type,
String<'.'>,
NumberToString<bugfix>::type
>::type;
constexpr const char* version_string = VersionString::value;
int main()
{
std::cout << version_string << std::endl;
}
See live example.
Here is my quick and dirty solution: http://coliru.stacked-crooked.com/a/43c9b365f6435991
It exploits the fact that a 'major.minor.fix' string will be quite short, so I just use a fixed-size array big enough to hold the string. The int-to-string function uses push_front
to prepend characters to string, so the whole string grows from end of buffer.
#include <cstdint>
constexpr const std::uint8_t major = 1;
constexpr const std::uint8_t minor = 10;
constexpr const std::uint8_t bugfix = 0;
struct char_array {
constexpr char_array() : ofs{sizeof(value) - 1}, value{} {}
constexpr const char* c_str() const { return value + ofs; }
constexpr void push_front(char c) {
--ofs;
value[ofs] = c;
}
private:
int ofs;
char value[42]; // big enough to hold version string
};
constexpr char_array predend_int(char_array a, int x) {
do {
auto digit = x % 10;
x = x / 10;
a.push_front(digit + '0');
}
while (x);
return a;
}
constexpr auto version_string() {
char_array a;
a = predend_int(a, bugfix);
a.push_front('.');
a = predend_int(a, minor);
a.push_front('.');
a = predend_int(a, major);
return a;
}
#include <iostream>
int main() {
constexpr char_array ver = version_string();
std::cout << ver.c_str() << '\n';
}
Update:
It's probably better to make a dedicated class for version string generation, and put all the functions in it: http://coliru.stacked-crooked.com/a/5e5ee49121cf6205
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