问题描述

我有一个类模板，用于获取变量的大小:

I have a class template which I use to get the size of a variable:

template <class T>
class Size
{
   unsigned int operator() (T) {return sizeof(T);}
};

这很好用，但对于字符串我想使用 strlen 而不是 sizeof:

This works fine but for strings I want to use strlen instead of sizeof:

template <>
class Size<char *>
{
   unsigned int operator() (char *str) {return strlen(str);}
};

问题是当我用 const char * 创建一个 size 实例时，它会转到非专业化版本.我想知道是否有办法在模板特化中同时捕获 char * 的常量和非常量版本?谢谢.

The problem is when I create an instance of size with const char * it goes to the unspecialized version. I was wondering if there is a way to capture both the const and non-const versions of char * in on template specialization? Thanks.

推荐答案

使用这个技巧:

#include <type_traits>

template< typename T, typename = void >
class Size
{
  unsigned int operator() (T) {return sizeof(T);}
};

template< typename T >
class Size< T, typename std::enable_if<
                 std::is_same< T, char* >::value ||
                 std::is_same< T, const char* >::value
               >::type >
{
  unsigned int operator() ( T str ) { /* your code here */ }
};

如何在类定义之外定义方法的示例.

Example of how to define the methods outside of the class definition.

添加了帮助程序以避免重复可能的冗长和复杂的条件.

Added helper to avoid repeating the possibly long and complex condition.

简化的助手.

#include <type_traits>
#include <iostream>

template< typename T >
struct my_condition
  : std::enable_if< std::is_same< T, char* >::value ||
                    std::is_same< T, const char* >::value >
{};

template< typename T, typename = void >
struct Size
{
  unsigned int operator() (T);
};

template< typename T >
struct Size< T, typename my_condition< T >::type >
{
  unsigned int operator() (T);
};

template< typename T, typename Dummy >
unsigned int Size< T, Dummy >::operator() (T)
{
  return 1;
}

template< typename T >
unsigned int Size< T, typename my_condition< T >::type >::operator() (T)
{
  return 2;
}

int main()
{
  std::cout << Size< int >()(0) << std::endl;
  std::cout << Size< char* >()(0) << std::endl;
  std::cout << Size< const char* >()(0) << std::endl;
}

哪个打印

1
2
2

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