Concatenate compile-time strings in a template at compile time?

不打扰是莪最后的温柔 提交于 2019-11-28 19:57:26

You could use something like this. Everything happens at compile time. Specialize base_typename_struct to define your primitive types.

template <const char* str, int len, char... suffix>
struct append {
  static constexpr const char* value() {
    return append<str, len-1, str[len-1], suffix...>::value();
  }
};

template <const char* str, char... suffix>
struct append<str, 0, suffix...> {
  static const char value_str[];
  static constexpr const char* value() {
    return value_str;
  }
};

template <const char* str, char... suffix>
const char append<str, 0, suffix...>::value_str[] = { suffix..., 0 };


template <typename T>
struct base_typename_struct;

template <>
struct base_typename_struct<int> {
  static constexpr const char name[] = "int";    
};


template <typename T, char... suffix>
struct typename_struct {
  typedef base_typename_struct<T> base;
  static const char* name() {
    return append<base::name, sizeof(base::name)-1, suffix...>::value();
  }
};

template <typename T, char... suffix>
struct typename_struct<T*, suffix...>:
  public typename_struct<T, '*', suffix...> {
};


int main() {
  cout << typename_struct<int****>::name() << endl;
}
Hedede

Alternative way without using recursive templates (but requires C++14):

#include <utility>
template<int...I> using is      = std::integer_sequence<int,I...>;
template<int N>   using make_is = std::make_integer_sequence<int,N>;

constexpr auto size(const char*s) { int i = 0; while(*s!=0){++i;++s;} return i; }

template<const char*, typename, const char*, typename>
struct concat_impl;

template<const char* S1, int... I1, const char* S2, int... I2>
struct concat_impl<S1, is<I1...>, S2, is<I2...>> {
    static constexpr const char value[]
    {
        S1[I1]..., S2[I2]..., 0
    };
};

template<const char* S1, const char* S2>
constexpr auto concat {
    concat_impl<S1, make_is<size(S1)>, S2, make_is<size(S2)>>::value
};

Example:

constexpr const char a[] = "int";
constexpr const char c[] = "**";

#include <iostream>
int main()
{
    std::cout << concat<a,b> << '\n';
}

append characters to string can also be implemented like this, by replacing the second const char* parameter with char....

Marco A.

I'm not sure of what you're searching for but I believe you're interested in a combination of typeid and name-demangling (which compiler are you using?)

In gcc it would be something like

#include<iostream>
#include <string>
#include <typeinfo>
#include <cstdlib>
#include <memory>
#include <cxxabi.h>
using namespace std;

std::string demangle(const char* name) {
    int status = -4; // some arbitrary value to eliminate the compiler warning

    // enable c++11 by passing the flag -std=c++11 to g++
    std::unique_ptr<char, void(*)(void*)> res {
        abi::__cxa_demangle(name, NULL, NULL, &status),
        std::free
    };
    return (status==0) ? res.get() : name ;
}

template <typename T> struct typename_struct {
  static std::string name() {
    std::string typeName = typeid(T).name();
    return demangle(typeName.c_str());
  }
};

int main(){

  cout << typename_struct<int****>::name(); // Prints "int****"

  return 0;
}

http://ideone.com/nLsFF0

Sources: https://stackoverflow.com/a/4541470/1938163

As for your question: those aren't constexpr constructs, thus the evaluation happens at runtime although the templated parameters and code are instantiated at compile-time.

Using templates doesn't mean every instruction contained in there will be executed and resolved at "compile-time".

I believe you can't achieve this whole bunch of stuff at compile-time since there are de-mangling functions (ABI-specific) involved. If I interpreted your question wrong please let me know.

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