如何简化升压变体的加号动作？

Question

我有 boost 变体类型定义：

typedef boost::variant< bool, int, float, double> VariantType;

我想对其执行 add/subtract/multiply/divide 操作。以添加 class 为例。问题是如果将新类型添加到 VariantType，例如 std::string，必须使用新类型更新添加 class。

struct Add : public boost::static_visitor<VariantType> {
    template <typename T>
    T operator() (T a, T b) const {
        return a + b;
    }
    float operator() (int a, float b) const {
        return a + b;
    }
    float operator() (float a, int b) const {
        return a + b;
    }
    double operator() (int a, double b) const {
        return a + b;
    }
    double operator() (double a, int b) const {
        return a + b;
    }
    double operator() (float a, double b) const {
        return a + b;
    }
    double operator() (double a, float b) const {
        return a + b;
    }
    VariantType operator() (bool a, int b) const {
        throw std::invalid_argument("bool and int can't Plus");
    }
    VariantType operator() (bool a, float b) const {
        throw std::invalid_argument("bool and float can't Plus");
    }
    VariantType operator() (bool a, double b) const {
        throw std::invalid_argument("bool and double can't Plus");
    }
    VariantType operator() (int a, bool b) const {
        throw std::invalid_argument("int and bool can't Plus");
    }
    VariantType operator() (float a, bool b) const {
        throw std::invalid_argument("float and bool can't Plus");
    }
    VariantType operator() (double a, bool b) const {
        throw std::invalid_argument("double and bool can't Plus");
    }
};

用法：

VariantType v1 = 1;
VariantType v2 = 2.1;
VariantType v3 = boost::apply_visitor(Add(), v1, v2);
cout<<boost::get<double>(v3)<<endl; //Print 2.2

GCC版本为4.8.2； boost版本为1.57.0；如何简单地添加class？谢谢

Answer 1

就让它成为一个多态仿函数：

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#include <boost/variant.hpp>
#include <iostream>

using VariantType = boost::variant<int, float, double, bool>;

struct Add : public boost::static_visitor<VariantType> {
    template <typename T, typename U>
    auto operator() (T a, U b) const -> decltype(a+b) { return a + b; }

    template <typename T> VariantType operator()(bool, T) const { throw std::invalid_argument("Can't to bool"); }
    template <typename T> VariantType operator()(T, bool) const { throw std::invalid_argument("Can't add bool"); }
    VariantType                     operator()(bool,bool) const { throw std::invalid_argument("Can't add bools"); }
};

int main() {
    std::cout << std::boolalpha;

    VariantType specimens[] = { int(42), 3.14f, 3.14, true };
    for (auto lhs : specimens)
        for (auto rhs : specimens)
        {
            try {
                std::cout << lhs << " + " << rhs << " ==  " << boost::apply_visitor(Add{}, lhs, rhs) << "\n";
            } catch(std::exception const& e) {
                std::cout << lhs << " + " << rhs << " ==> " << e.what() << "\n";
            }
        }
}

打印：

42 + 42 ==  84
42 + 3.14 ==  45.14
42 + 3.14 ==  45.14
42 + true ==> Can't add bool
3.14 + 42 ==  45.14
3.14 + 3.14 ==  6.28
3.14 + 3.14 ==  6.28
3.14 + true ==> Can't add bool
3.14 + 42 ==  45.14
3.14 + 3.14 ==  6.28
3.14 + 3.14 ==  6.28
3.14 + true ==> Can't add bool
true + 42 ==> Can't to bool
true + 3.14 ==> Can't to bool
true + 3.14 ==> Can't to bool
true + true ==> Can't add bools

如何简化升压变体的加号动作？

How to simplify the plus action on boost variant?

c++

boost

boost-variant