不带任意参数的c ++函数模板 [英] c++ function templates with not arbitrary parameters

问题描述

考虑这个工作正常的c ++代码：

  template< size_t N& 
 int check（std :: array< unsigned char，N> buf）{
 std :: cout< N<  - << buf.size（）<< std :: endl; 
 return 0; 
} 
 
 
 int main（int argc，char * argv []）
 {
 std :: array< unsigned char，20>一个; 
 std :: array< unsigned char，30> b; 
 std :: array< unsigned char，40> C; 
 
 check（a）; 
 check（b）; 
 check（c）; 
 
 return 0; 
} 
  

可以显式实例化检查N = 20和N = 30，但是禁用任何其他隐式实例化？

这意味着如果我使用check（c）

编辑：

事实上我想有相同的实现，所以一个模板是正确的选择，我想只是有一个编译时错误，如果由于某种原因我实例化它的参数不应该存在。
因为static_assert（c ++ 11）解决方案看起来是最好的，完全的问题，非常人类可读。

N ，则可以使用 static_assert 引发编译时错误。是除20或30之外的任何值。

  template< size_t N> 
 int check（std :: array< unsigned char，N> buf）
 {
 static_assert（N == 20 || N == 30，N必须是20或30 ）; 
 
 std :: cout<< N<  - << buf.size（）<< std :: endl; 
 
 return 0; 
} 
  

Consider this c++ code that works fine:

template <size_t N>
int check(std::array<unsigned char, N> buf){
    std::cout << N << "-" << buf.size() << std::endl;
    return 0;
}


int main (int argc, char *argv[])
{
    std::array<unsigned char, 20> a;
    std::array<unsigned char, 30> b;
    std::array<unsigned char, 40> c;

    check(a);
    check(b);
    check(c);

    return 0;
}

Is it possible to explicitly instantiate "check" for N=20 and N=30, but disable any other implicit instantiation?

That means that I would like to get a compile time error if I use "check(c)",

Edit:

In fact I wanted to have the same implementation, so a template was the right choice, and I wanted to just have a compile time error if for some reason I instantiate it with a parameters that is not supposed to exist. Because of that the static_assert (c++11) solution looks the best, as complete for the problem and very human readable.

解决方案

You can use static_assert to throw a compile time error if N is anything other than 20 or 30.

template <size_t N>
int check(std::array<unsigned char, N> buf) 
{
    static_assert(N == 20 || N == 30, "N must be 20 or 30.");

    std::cout << N << "-" << buf.size() << std::endl;

    return 0;
}

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