字符串 class 构造函数接受 const char* 列表

Question

In my String class, I want to make a ctor that accept variable length of arguments ( const char* ) like parameter pack with variadic template.

Something like following

class String
{
public: 
    
    String(const char*... xList)
    {
         // Internally appending all const char* present in xList and 
         // populating String class char buffer
    } 

};

Any idea how to achieve this? I was under impression that following thing should work... but it is not.

template <const char* ... Args>
String(Args... xList)
{
    
} 

There may be a way by using va_arg but I am not interested in that.

Answer 1

The way to achieve what you want is to take arguments of arbitrary types and then to constrain them using SFINAE . In C++11/14, you'll need some additional boilerplate code:

template<class...>
struct conjunction : std::true_type {};

template<class B>
struct conjunction<B> : B {};

template<class B, class... Bn>
struct conjunction<B, Bn...> 
    : std::conditional_t<B::value, conjunction<Bn...>, B> {};

class String
{
public: 
    template<class... Args, typename = std::enable_if_t<
        conjunction<std::is_convertible<Args, const char*>...>::value>>
    String(const Args&... list)
    {} 
};

In C++17, the solution is shorter:

class String
{
public: 
    template<class... Args, typename = std::enable_if_t<
        (std::is_convertible_v<Args, const char*> && ...)>>
    String(const Args&... list)
    {} 
};

Answer 2

Here is a working example using variadic templates and SFINAE which won't compile if you don't provide const char* s

#include <type_traits>
#include <list>
#include <iostream>

struct String
{
    std::list<char> ls;
    template<class ... Args, class = std::enable_if_t<std::is_same_v<std::common_type_t<Args...>, const char *>>>
    String(Args...args):ls{*args...}
    {
    }

};

int main(){
    auto c1 = 'a'; const char * pc1 = &c1;
    auto c2 = 'b'; const char * pc2 = &c2;
    auto c3 = 'c'; const char * pc3 = &c3;
    String s{pc1, pc2, pc3};

    for(auto it = s.ls.begin();it != s.ls.end(); ++it)std::cout << *it;
}

Live

With c++11 , the code can be like this

#include <type_traits>
#include <list>
#include <iostream>

struct String
{
    std::list<char> ls;

    template<class ... Args, class = typename std::enable_if<std::is_same<typename std::common_type<Args...>::type, const char *>::value>::type>
    String(Args...args):ls{*args...}
    {
    }

};

int main(){
    auto c1 = 'a'; const char * pc1 = &c1;
    auto c2 = 'b'; const char * pc2 = &c2;
    auto c3 = 'c'; const char * pc3 = &c3;
    String s{pc1, pc2, pc3};

    for(auto it = s.ls.begin();it != s.ls.end(); ++it)std::cout << *it;
}

Live

Answer 3

Not exactly what you asked but... if you accept an additional couple of brackets calling the constructor, you can pass your variadic list of char const * as follows

struct foo
 {
   template <std::size_t Dim>
   foo (char const * const (&arr)[Dim])
    { /* do something with arr */ }
 };

The drawback is that you can't call it as follows

foo f("one", "two", "three");

but you have to add the brackets

// ...V.....................V
foo f({"one", "two", "three"});

The following is a full compiling example

#include <iostream>

struct foo
 {
   template <std::size_t Dim>
   foo (char const * const (&arr)[Dim])
    {
      for ( auto const & str : arr )
         std::cout << str << std::endl;
    } 
};

int main ()
 { 
   foo f({"one", "two", "three"});
 }