我对定义一个要继承的通用类感兴趣,该通用类基于枚举和该枚举到某些数据结构的映射而有所不同。
在下面的代码中,我具有从test_parent继承的test_child,该对象具有已实现的共享功能。我的计划是让许多类从类似parent_class的类继承,但要定义一个唯一的“字段”枚举和相应的“映射”映射。
#include <iostream>
#include <string>
#include <unordered_map>
class test_parent {
public:
enum class field {
A,
B,
C
};
typedef struct {
std::string s;
int i, j;
} data_t;
std::unordered_map<field, data_t> mapping {
{field::A, {"A", 1, 1}},
{field::B, {"B", 2, 2}},
{field::C, {"C", 3, 3}}
};
int get_i (field f) {
return mapping[f].i;
}
std::string get_s (field f) {
return mapping[f].s;
}
};
class test_child : test_parent {
public:
enum class field {
D,
E
};
std::unordered_map<field, data_t> mapping {
{field::D, {"D", 4, 4}},
{field::E, {"E", 5, 5}}
};
};
int main () {
test_parent tp;
test_child tc;
std::cout << tp.get_i(test_parent::field::A) << " " << tc.get_i(test_child::field::E) << std::endl;
return 0;
}
此代码返回编译错误:
test.cpp: In function ‘int main()’:
test.cpp:55:86: error: no matching function for call to ‘test_child::get_i(test_child::field)’
std::cout << tp.get_i(test_parent::field::A) << " " << tc.get_i(test_child::field::E) << std::endl;
^
test.cpp:28:6: note: candidate: int test_parent::get_i(test_parent::field)
int get_i (field f) {
^~~~~
test.cpp:28:6: note: no known conversion for argument 1 from ‘test_child::field’ to ‘test_parent::field’
但是我希望打印的是:
1 5
答案 0 :(得分:0)
不确定是不是想要的,但是可以使用模板
struct data_t
{
std::string s;
int i;
int j;
};
template <typename E>
class test_parent {
public:
int get_i(E e) const { return mapping.at(e).i; }
const std::string& get_s(E e) const { return mapping.at(e).s; }
static const std::unordered_map<E, data_t> mapping;
};
然后
enum class field_ABC{ A, B, C };
enum class field_DE{ D, E };
template <>
const std::unordered_map<field_ABC , data_t> test_parent<field_ABC >::mapping = {
{field_ABC::A, {"A", 1, 1}},
{field_ABC::B, {"B", 2, 2}},
{field_ABC::C, {"C", 3, 3}}
};
template <>
const std::unordered_map<field_DE, data_t> test_parent<field_DE>::mapping = {
{field_DE::D, {"D", 4, 4}},
{field_DE::E, {"E", 5, 5}}
};
答案 1 :(得分:0)
嗨,我想我可能在以下示例中找到了答案:
#include <iostream>
#include <string>
#include <unordered_map>
class test_parent {
public:
enum class field {
A,
B,
C
};
typedef struct {
std::string s;
int i, j;
} data_t;
std::unordered_map<uint, data_t> mapping {
{(uint) field::A, {"A", 1, 1}},
{(uint) field::B, {"B", 2, 2}},
{(uint) field::C, {"C", 3, 3}}
};
int get_i (uint f) {
return mapping[f].i;
}
std::string get_s (uint f) {
return mapping[f].s;
}
};
class test_child : public test_parent {
public:
test_child () {
test_parent::mapping = mapping;
}
enum class field {
D = 4,
E = 5
};
std::unordered_map<uint, data_t> mapping {
{(uint) field::E, {"E", 5, 5}},
{(uint) field::D, {"D", 4, 4}}
};
};
int main () {
test_parent tp;
test_child tc;
std::cout << tp.get_i((uint) test_parent::field::A) << " " << tc.get_i((uint) test_child::field::E) << std::endl;
return 0;
}
在我将枚举值最初视为整数的情况下,它是整数!