有一篇关于 C++ 用于科学计算的精彩论文，其中作者 (T. Veldhuizen) 提出了一种基于特征的方法来解决类型提升问题。我使用过这种方法，发现它很有效：
#include<iostream>
#include<complex>
#include<typeinfo>
template<typename T1, typename T2>
struct promote_trait{};
#define DECLARE_PROMOTION(A, B, C) template<> struct promote_trait<A, B> { using T_promote = C;};
DECLARE_PROMOTION(int, char, int);
DECLARE_PROMOTION(int, float, float);
DECLARE_PROMOTION(float, std::complex<float>, std::complex<float>);
// similarly for all possible type combinations...
template<typename T1, typename T2>
void product(T1 a, T2 b) {
using T = typename promote_trait<T1, T2>::T_promote;
T ans = T(a) * T(b);
std::cout<<"received "
<<typeid(T1).name()<<"("<<a<<")"<<" * "
<<typeid(T2).name()<<"("<<b<<")"<<" ==> "
<<"returning "
<<typeid(T).name()<<"("<<ans<<")"<<std::endl;
}
int main() {
product(1, 'a');
product(1, 2.0f);
product(1.0f, std::complex<float>(1.0f, 2.0f));
return 0;
}
输出：
received i(1) * c(a) ==> returning i(97)
received i(1) * f(2) ==> returning f(2)
received f(1) * St7complexIfE((1,2)) ==> returning St7complexIfE((1,2))
typeinfo 返回的类型名称取决于实现；您的输出可能与我的不同，后者在 OS X 10.7.4 上使用 GCC 4.7.2
本质上，该方法定义了一个promote_trait只包含一个类型定义的类型：当以给定方式操作时，两种类型应该提升到的类型。需要声明所有可能的促销活动。
当一个函数同时接收这两种类型时，它依赖于promote_trait推断出正确的、提升的结果类型。如果没有为给定的对定义特征，则代码无法编译（一个理想的特性）。
现在，有问题的论文写于 2000 年，我们知道 C++ 在过去十年中发生了巨大的变化。那么，我的问题如下：
是否有一种现代的、惯用的 C++ 11 方法来处理类型提升，与 Veldhuizen 引入的基于特征的方法一样有效？
编辑（关于使用std::common_type）
根据 Luc Danton 的建议，我创建了以下代码，它使用std::common_type：
#include<iostream>
#include<complex>
#include<typeinfo>
#include<typeindex>
#include<string>
#include<utility>
#include<map>
// a map to homogenize the type names across platforms
std::map<std::type_index, std::string> type_names = {
{typeid(char) , "char"},
{typeid(int) , "int"},
{typeid(float) , "float"},
{typeid(double) , "double"},
{typeid(std::complex<int>) , "complex<int>"},
{typeid(std::complex<float>) , "complex<float>"},
{typeid(std::complex<double>) , "complex<double>"},
};
template<typename T1, typename T2>
void promotion(T1 a, T2 b) {
std::string T1name = type_names[typeid(T1)];
std::string T2name = type_names[typeid(T2)];
std::string TPname = type_names[typeid(typename std::common_type<T1, T2>::type)];
std::cout<<T1name<<"("<<a<<") and "<<T2name<<"("<<b<<") promoted to "<<TPname<<std::endl;
}
int main() {
promotion(1, 'a');
promotion(1, 1.0);
promotion(1.0, 1);
promotion(std::complex<double>(1), 1);
promotion(1.0f, 1);
promotion(1.0f, 1.0);
promotion(std::complex<int>(1), std::complex<double>(1));
promotion(std::complex<double>(1), std::complex<int>(1));
promotion(std::complex<float>(0, 2.0f), std::complex<int>(1));
return 0;
}
输出：
int(1) and char(a) promoted to int
int(1) and double(1) promoted to double
double(1) and int(1) promoted to double
complex<double>((1,0)) and int(1) promoted to complex<double>
float(1) and int(1) promoted to float
float(1) and double(1) promoted to double
complex<int>((1,0)) and complex<double>((1,0)) promoted to complex<int>
complex<double>((1,0)) and complex<int>((1,0)) promoted to complex<int>
complex<float>((0,2)) and complex<int>((1,0)) promoted to complex<int>
我惊讶地发现，除了最后三个促销活动之外，所有促销活动都符合我的预期。为什么会complex<int>和complex<double>或complex<float>被提升为complex<int>！？