hw03 Gelee #18
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hw03 Gelee #18
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archibate
approved these changes
Jan 1, 2022
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| template <class T1, class T2> | ||
| auto operator+(std::variant<T1, T2> const &a, T2 const &b) { | ||
| std::variant<T1, T2> b1 = b; |
| // } | ||
| //}; | ||
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| template <class T1, class T2> | ||
| std::ostream &operator<<(std::ostream &os, std::variant<T1, T2> const &a) { | ||
| // 请实现自动匹配容器中具体类型的打印！10 分 |
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| template <class T1, class T2> | ||
| std::ostream &operator<<(std::ostream &os, std::variant<T1, T2> const &a) { | ||
| // 请实现自动匹配容器中具体类型的打印！10 分 | ||
| std::visit([&](auto const & it){ | ||
| os << it << std::endl; |
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这里不需要 endl 哦！一般调用该函数时候 cout << d << endl 已经有一个了。
| } | ||
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| template <class T1, class T2> | ||
| std::variant<T1, T2> operator+(std::variant<T1, T2> const &a, std::variant<T1, T2> const &b) { | ||
| // 请实现自动匹配容器中具体类型的加法！10 分 | ||
| } | ||
| return std::visit([&] (auto const &t1, auto const &t2) | ||
| ->std::variant<T1 ,T2>{ |
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做的好！直接在返回类型上构造，避免了额外的默认构造+拷贝赋值。
| for(int i =0; i < n ; i++){ | ||
| res[i] = a[i]+b[i]; | ||
| } | ||
| return res; | ||
| } | ||
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| template <class T1, class T2> | ||
| std::variant<T1, T2> operator+(std::variant<T1, T2> const &a, std::variant<T1, T2> const &b) { | ||
| // 请实现自动匹配容器中具体类型的加法！10 分 |
| // 请实现列表的逐元素加法！10 分 | ||
| // 例如 {1, 2} + {3, 4} = {4, 6} | ||
| using T0 = decltype(T1{} + T2{}); | ||
| const int n = std::min(a.size(),b.size()); | ||
| std::vector<T0> res(n,0); |
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不需要指定右边的 0，反而是如果 T0 不是一个数值类型这样会失败，直接 std::vector(n) 就可以调用数值类型 T0 的默认的构造函数 - 零初始化了。
| @@ -16,19 +17,68 @@ std::ostream &operator<<(std::ostream &os, std::vector<T> const &a) { | |||
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| // 请修复这个函数的定义：10 分 | |||
| @@ -3,6 +3,7 @@ | |||
| #include <variant> | |||
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| // 请修复这个函数的定义：10 分 | |||
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修复第一个函数：简单的加上template即可；
修复第二个函数：由下面的实际运算结果可知，结果的vector与二者中最短的相同，逐项相加即可。
自动匹配容器中具体类型的方法：
此“+” 为 std::variant<T1, T2> + std::variant<T1, T2>; 使用课上讲的std::visit，有多个variant作为参数，visit会自动地罗列出所
有的排列组合。
自动匹配容器中具体类型的打印: 同理的使用std::visit即可。
关于
d + c + e:编译器给的报错是：
由此有了两种思路：
对于
d + c，重载运算符+，可以将c变为d的类型后，调用上面已写好的std::variant<T1, T2> + std::variant<T1, T2>的重载运算符+即可，这样子其实d + c + e 已经可以运算。但是若有表达式d = c + e， 则是不可以的。为了防止未知的情况，所以对于c在运算符的左边的情况也需要重载。第二种思路则是利用
std::hold_alternative或者v.index来获取std::variant<T1, T2>类型，然后对于d + c, c + e重载运算符将+时，使用std::get<index>()将d ，e变为确定的类型， 然后调用关于std::vector<T1> + std::vector<T2>的重载运算符“+”即可。