C++98 从运算符获取提升结果

Question

Type traits to get result of promotion from operator

关于此的后续问题是否有仅使用 C++98 且不提升的好方法？

Answer 1

只要你只需要担心标准的内置类型，你就可以这样做:

template <int> struct plus_helper;
template <> struct plus_helper< 1> { typedef char type; };
template <> struct plus_helper< 2> { typedef signed char type; };
template <> struct plus_helper< 3> { typedef unsigned char type; };
template <> struct plus_helper< 4> { typedef short type; };
template <> struct plus_helper< 5> { typedef unsigned short type; };
template <> struct plus_helper< 6> { typedef int type; };
template <> struct plus_helper< 7> { typedef unsigned int type; };
template <> struct plus_helper< 8> { typedef long type; };
template <> struct plus_helper< 9> { typedef unsigned long type; };
template <> struct plus_helper<10> { typedef float type; };
template <> struct plus_helper<11> { typedef double type; };
template <> struct plus_helper<12> { typedef long double type; };
template <> struct plus_helper<13> { typedef wchar_t type; };

template <typename T1, typename T2>
struct plus {
private:
  static char (&f(char))[1];
  static char (&f(signed char))[2];
  static char (&f(unsigned char))[3];
  static char (&f(short))[4];
  static char (&f(unsigned short))[5];
  static char (&f(int))[6];
  static char (&f(unsigned int))[7];
  static char (&f(long))[8];
  static char (&f(unsigned long))[9];
  static char (&f(float))[10];
  static char (&f(double))[11];
  static char (&f(long double))[12];
  static char (&f(wchar_t))[13];
public:
  typedef typename plus_helper<sizeof(f(*(T1*)0 + *(T2*)0))>::type type;
};

template <typename T1, typename T2>
struct plus<T1 *, T2> {
  typedef T1 *type;
};

template <typename T1, typename T2>
struct plus<T1, T2 *> {
  typedef T2 *type;
};

请注意，我要求可以添加 T1 和 T2。如果不能，则不一定给出错误。另请注意，我没有省略短于 int 的类型。它们永远不会由内置的 + 运算符返回，但它们是自定义 operator+ 的有效返回类型。

不幸的是，如果您还需要担心其他类型，例如自定义结构，这种方法将行不通，虽然我很乐意被证明是错误的，但我认为不可能做到这一点工作。