Passing a string literal as a type argument to a class template

Question

I want to declare a class template in which one of the template parameters takes a string literal, e.g. my_class<"string">.

Can anyone give me some compilable code which declares a simple class template as described?

---

Note: The previous wording of this question was rather ambiguous as to what the asker was actually trying to accomplish, and should probably have been closed as insufficiently clear. However, since then this question became multiple times referred-to as the canonical ‘string literal type parameter’ question. As such, it has been re-worded to agree with that premise.

Answer 1

You can have a const char* non-type template parameter, and pass it a const char[] variable with static linkage, which is not all that far from passing a string literal directly.

#include <iostream>    

template<const char *str> 
struct cts {
    void p() {std::cout << str;}
};

static const char teststr[] = "Hello world!";
int main() {
    cts<teststr> o;
    o.p();
}

http://coliru.stacked-crooked.com/a/64cd254136dd0272

Answer 2

Further from Neil's answer: one way to using strings with templates as you want is to define a traits class and define the string as a trait of the type.

#include <iostream>

template <class T>
struct MyTypeTraits
{
   static const char* name;
};

template <class T>
const char* MyTypeTraits<T>::name = "Hello";

template <>
struct MyTypeTraits<int>
{
   static const char* name;
};

const char* MyTypeTraits<int>::name = "Hello int";

template <class T>
class MyTemplateClass
{
    public:
     void print() {
         std::cout << "My name is: " << MyTypeTraits<T>::name << std::endl;
     }
};

int main()
{
     MyTemplateClass<int>().print();
     MyTemplateClass<char>().print();
}

prints

My name is: Hello int
My name is: Hello

Answer 3

C++20 fixed_string + "Class Types in Non-Type Template Parameters"

Apparently, a proposal for this was first accepted, but then removed: "String literals as non-type template parameters"

The removal was partly because it was deemed to be easy enough to do with another proposal that was accepted: "Class Types in Non-Type Template Parameters".

The accepted proposal contains an example with the following syntax:

template <std::basic_fixed_string Str>
struct A {};
using hello_A = A<"hello">;

I'll try to update this with an example that actually tells me anything once I see a compiler that supports it.

A Redditor has also shown that the following compiles on GCC master, provided you define your own version of basic_fixed_string which was not in the standard library yet: https://godbolt.org/z/L0J2K2

template<unsigned N>
struct FixedString {
    char buf[N + 1]{};
    constexpr FixedString(char const* s) {
        for (unsigned i = 0; i != N; ++i) buf[i] = s[i];
    }
    constexpr operator char const*() const { return buf; }
};
template<unsigned N> FixedString(char const (&)[N]) -> FixedString<N - 1>;

template<FixedString T>
class Foo {
    static constexpr char const* Name = T;
public:
    void hello() const;
};

int main() {
    Foo<"Hello!"> foo;
    foo.hello();
}

g++ -std=c++2a 9.2.1 from the Ubuntu PPA fails to compile that with:

/tmp/ccZPAqRi.o: In function `main':
main.cpp:(.text+0x1f): undefined reference to `_ZNK3FooIXtl11FixedStringILj6EEtlA7_cLc72ELc101ELc108ELc108ELc111ELc33EEEEE5helloEv'
collect2: error: ld returned 1 exit status

Bibliography: https://botondballo.wordpress.com/2018/03/28/trip-report-c-standards-meeting-in-jacksonville-march-2018/

Finally, EWG decided to pull the previously-approved proposal to allow string literals in non-type template parameters, because the more general facility to allow class types in non-type template parameters (which was just approved) is a good enough replacement. (This is a change from the last meeting, when it seemed like we would want both.) The main difference is that you now have to wrap your character array into a struct (think fixed_string or similar), and use that as your template parameter type. (The user-defined literal part of P0424 is still going forward, with a corresponding adjustment to the allowed template parameter types.)

This will be especially cool with the C++17 if constexpr: if / else at compile time in C++?

This kind of feature appears to be in line with the awesome "constexpr everything" proposals that went into C++20, such as: Is it possible to use std::string in a constexpr?

Answer 4

Sorry, C++ does not currently support the use of string literals (or real literals) as template parameters.

But re-reading your question, is that what you are asking? You cannot say:

foo <"bar"> x;

but you can say

template <typename T>
struct foo {
   foo( T t ) {}
};

foo <const char *> f( "bar" );

Answer 5

This is a solution with MPLLIBS to pass a strings as template arguments ( C++11 ).

#include <iostream>
#include <mpllibs/metaparse/string.hpp> // https://github.com/sabel83/mpllibs
#include <boost/mpl/string.hpp>

// -std=c++11

template<class a_mpl_string>
struct A
{
  static const char* string;
};

template<class a_mpl_string>
const char* A< a_mpl_string >
::string { boost::mpl::c_str< a_mpl_string >::value };  // boost compatible

typedef A< MPLLIBS_STRING ( "any string as template argument" ) > a_string_type;

int main ( int argc, char **argv )
{
  std::cout << a_string_type{}.string << std::endl;
  return 0;
}

prints:

any string as template argument

The lib on github: https://github.com/sabel83/mpllibs

Answer 6

inline const wchar_t *GetTheStringYouWant() { return L"The String You Want"; }

template <const wchar_t *GetLiteralFunc(void)>
class MyType
{
     void test()
     {
           std::cout << GetLiteralFunc;
     }    
}

int main()
{
     MyType<GetTheStringYouWant>.test();
}

Try it with pasing the address of a function as the template argument.

Answer 7

EDIT: ok the title of your question seems to be misleading

"I want a class which takes two parameters in its constructor. The first can be either an int, double or float, so , and the second is always a string literal "my string", so I guess const char * const."

It looks like you're trying to achieve:

template<typename T>
class Foo
{
  public:
  Foo(T t,  const char* s) : first(t), second(s)
  {
    // do something
  }

  private:
  T first;
  const char* second;

};

This would work for any type, for the first parameter: int, float, double, whatever.

Now if you really want to restrict the type of the first parameter to be only int, float or double; you can come up with something more elaborate like

template<typename T>
struct RestrictType;

template<>
struct RestrictType<int>
{
  typedef int Type;
};

template<>
struct RestrictType<float>
{
  typedef float Type;
};

template<>
struct RestrictType<double>
{
  typedef double Type;
};

template<typename T>
class Foo
{
  typedef typename RestrictType<T>::Type FirstType;

  public:
  Foo(FirstType t,  const char* s) : first(t), second(s)
  {
    // do something
  }

  private:
  FirstType first;
  const char* second;

};

int main()
{
  Foo<int> f1(0, "can");
  Foo<float> f2(1, "i");
  Foo<double> f3(1, "have");
  //Foo<char> f4(0, "a pony?");
}

If you remove the comment on the last line, you'll effectively get a compiler error.

---

String literals are not allowed by C++2003

ISO/IEC 14882-2003 §14.1:

14.1 Template parameters

A non-type template-parameter shall have one of the following (optionallycv-qualified) types:

— integral or enumeration type,

— pointer to object or pointer to function,

— reference to object or reference to function,

— pointer to member.

ISO/IEC 14882-2003 §14.3.2:

14.3.2 Template non-type arguments

A template-argument for a non-type, non-template template-parameter shall be one of:

— an integral constant-expression of integral or enumeration type; or

— the name of a non-type template-parameter; or

— the address of an object or function with external linkage, including function templates and function template-ids but excluding non-static class members, expressed as & id expression where the & is optional if the name refers to a function or array, or if the corresponding template-parameter is a reference; or

— a pointer to member expressed as described in 5.3.1.

[Note:A string literal (2.13.4) does not satisfy the requirements of any of these categories and thus is not an acceptable template-argument.

[Example:

template<class T, char* p> class X { 
  //... 
  X(); 
  X(const char* q) { /* ... */ } 
}; 

X<int,"Studebaker"> x1; //error: string literal as template-argument 
char p[] = "Vivisectionist"; 
X<int,p> x2; //OK 

—end example] —end note]

And it looks like it's not going to change in the upcoming C++0X, see the current draft 14.4.2 Template non-type arguments.

Answer 8

Based on your comments under Niel's answer, another possibility is the following:

#include <iostream>

static const char* eventNames[] = { "event_A", "event_B" };

enum EventId {
        event_A = 0,
        event_B
};

template <int EventId>
class Event
{
public:
   Event() {
     name_ = eventNames[EventId];
   }
   void print() {
        std::cout << name_ << std::endl;
   }
private:
   const char* name_;
};

int main()
{
        Event<event_A>().print();
        Event<event_B>().print();
}

prints

event_A
event_B

Answer 9

You cannot pass a string literal directly as a template parameter.

But you can get close:

template<class MyString = typestring_is("Hello!")>
void MyPrint() {
  puts( MyString::data() );
}

...
// or:
MyPrint<typestring_is("another text")>();
...

All you need is a small header file from here.

---

Alternatives:

• Define a global char const * and pass it to the template as pointer. (here)

  Drawback: Requires additional code outside of the template argument list. It is not suitable, if you need to specify the string literal "inline".

• Use a non-standard language extension. (here)

  Drawback: Not guaranteed to work with all compilers.

• Use BOOST_METAPARSE_STRING. (here)

  Drawback: Your code will depend on the Boost library.

• Use a variadic template parameter pack of char, e.g. str_t<'T','e','s','t'>.

  This is what the above solution does for you behind the scenes.

Answer 10

Use proxy static constexpr const char type_name_str[] = {"type name"}; for passing string as template parameter. Defining string using [] is important.

#include <iostream>

template<typename T, const char* const t_name>
struct TypeName
{
public:

    static constexpr const char* Name()         
    {                                   
        return t_name;
    };                                  

};

static constexpr const char type_name_str[] = {"type name"};

int main() 
{
    std::cout<<TypeName<float, type_name_str>::Name();
    return 0;
}

Answer 11

I want a class which takes two parameters in its constructor. The first can be either an int, double or float, so , and the second is always a string literal "my string"

template<typename T>
class demo
{
   T data;
   std::string s;

   public:

   demo(T d,std::string x="my string"):data(d),s(x) //Your constructor
   {
   }
};

I am not sure but is this something what you want?

Answer 12

I was struggling with a similar problem and finally came up with a concise implementation that unpacks the string literal into a char... template parameter pack and without using the GNU literal operator template extension:

#include <utility>

template <char ...Chars>
struct type_string_t {
    static constexpr const char data[sizeof...(Chars)] = {Chars...};
};

template <char s(std::size_t), std::size_t ...I>
auto type_string_impl(std::index_sequence<I...>) {
    return type_string_t<s(I)...>();
}

#define type_string(s) \
    decltype (type_string_impl<[] -> constexpr (std::size_t i) {return s[i];}> \
        (std::make_index_sequence<sizeof (s)>()))

static_assert (std::is_same<type_string("String_A"),
                            type_string("String_A")>::value);
static_assert (!std::is_same<type_string("String_A"),
                             type_string("String_B")>::value);

A major caveat: this depends on a C++20 feature (class values as non-type template arguments; P0732, P1907), which (as of December 2020) is only (partially) implemented in GCC 9 and later (preprocessor feature test: (__cpp_nontype_template_args >= 201911L) || (__GNUG__ >= 9)). However, since the feature is standard, it is only a matter of time before other compilers catch up.

Answer 13

Maybe not what the OP is asking, but if you use boost, you can create a macro like this for example:

#define C_STR(str_) boost::mpl::c_str< BOOST_METAPARSE_STRING(str_) >::value

Then use as follows:

template<const char* str>
structe testit{
};
testit<C_STR("hello")> ti;

Answer 14

template <char... elements>
struct KSym /* : optional_common_base */ {
  // We really only care that we have a unique-type and thus can exploit being a `""_ksym singleton`
  const char z[sizeof...(elements) + 1] = { elements..., '\0' };
  // We can have properties, we don't need anything to be constexpr for Rs
};
template <typename T, T... chars>
auto&& operator""_ksym() { 
  static KSym<chars...> kSym; // Construct the unique singleton (lazily on demand)
  return kSym;
}
static auto ksym_example1 = "a unique string symbol1\n"_ksym.z;
static auto ksym_example2 = "a unique string symbol2\n"_ksym.z;
auto dont_care = []() {
  ::OutputDebugString(ksym_example1);
  ::OutputDebugString("a unique string symbol2\n"_ksym.z);
  assert("a unique string symbol1\n"_ksym.z == ksym_example1);
  assert("a unique string symbol2\n"_ksym.z == ksym_example2);
  return true; 
}();

The above is working for me in production using Clang 11 on Windows.

(edited) I now use exactly this in clang on Windows:

// P0424R1: http://www.open-std.org/jtc1/SC22/wg21/docs/papers/2017/p0424r1.pdf
template <char... chars_ta> struct KSymT;
template <typename T, T... chars_ta> // std::move(KSymT<chars_ta...>::s);
auto operator""_ksym()->KSymT<chars_ta...>& { return KSymT<chars_ta...>::s; }
struct KSym {
  virtual void onRegister() {}
  virtual std::string_view zview_get() = 0;
};

template <char... chars_ta>
struct KSymT : KSym {
  inline static KSymT s;
  // We really only care that we have a unique-type and thus can exploit being a `""_ksym singleton`
  inline static constexpr char z[sizeof...(chars_ta) + 1] = { chars_ta..., '\0' };
  inline static constexpr UIntPk n = sizeof...(chars_ta);
  // We can have properties, we don't need anything to be constexpr for Rs
  virtual std::string_view zview_get() { return std::string_view(z); };
  //#KSym-support compare with `Af_CmdArgs`
  inline bool operator==(const Af_CmdArgs& cmd) {
    return (cmd.argl[0] == n && memcmp(cmd.argv[0], z, n) == 0);
  }
};

Answer 15

Another C++20 solution I don't see mentioned, but which was sufficiently simple and suitable for my own needs, is to use a constexpr lambda as the NTTP returning the string:

#include <string_view>

template<auto getStrLambda>
struct MyType {
    static constexpr std::string_view myString{getStrLambda()};
};

int main() {
    using TypeWithString = MyType<[]{return "Hello world!";}>;
    return 0;
}

Compiler explorer example here.

Answer 16

here is a solution and extensions/examples

my solution extends https://ctrpeach.io/posts/cpp20-string-literal-template-parameters/

#include <iostream>
#include <algorithm>
#include <string>

template<size_t N>
struct StringLiteral {
    char value[N];
    constexpr StringLiteral(const char(&str)[N]) {
        std::copy_n(str, N, value);
    }
};

template <StringLiteral T>
struct String {
    static constexpr std::string str() {
        return T.value;
    }
};

template <typename... Strings>
struct JoinedString {
    static constexpr std::string str() {
        return (Strings::str() + ...);
    }
};

template <typename Delim, typename String, typename... Strings>
struct DelimJoinedString {
    static constexpr std::string str() {
        if constexpr (sizeof...(Strings))
            return JoinedString<String, Delim, DelimJoinedString<Delim, Strings...>>::str();
        else
            return String::str();
    }
};

int main() {
    // "123"
    using s123 = String<"123">;
    std::cout << s123::str() << "\n";

    // "abc"
    using abc = String<"abc">;
    std::cout << abc::str() << "\n";

    // "abc123abc123"
    using abc123abc123 = JoinedString<abc, s123, abc, s123>;
    std::cout << abc123abc123::str() << "\n";

    // "abc, 123"
    using abccomma123 = DelimJoinedString<String<", ">, abc, s123>;
    std::cout << abccomma123::str() << "\n";

    // "abc, 123, 123, abc"
    using commaabc123123abc = DelimJoinedString<String<", ">, abc, s123, s123, abc>;
    std::cout << commaabc123123abc::str() << "\n";
    return 0;
}

Answer 17

a string literal "my string", so I guess const char * const

Actually, string literals with n visible characters are of type const char[n+1].

#include <iostream>
#include <typeinfo>

template<class T>
void test(const T& t)
{
    std::cout << typeid(t).name() << std::endl;
}

int main()
{
    test("hello world"); // prints A12_c on my compiler
}