C++:将未知 union 传递给函数作为参数

Question

所以我有这样的东西:

struct FoodType {
    union {
        int fat = 2;
    } Apple;
    union {
        int fat = 3;
    } Banana;
} FoodType;

我想编写一个函数，它接受一个“未知”的 FoodType 作为参数。

void eatFood(FoodType type) { /* do something */ }

如何实现？

Answer 1

您在问题中提供的枚举解决方案的替代方案是实例化:

#include <iostream>

struct FoodType {
    FoodType(int f) : fat(f) {}
    int getFat() { return fat; }
    int fat;
};

void eatFood(const FoodType& type) { std::cout << "I ate " << type.getFat() << " grams of fat\n"; }

int main() {
    FoodType apple(2);
    eatFood(apple);
    FoodType banana(3);
    eatFood(banana);
}

或者，对于更复杂的情况，可以使用多态性，但在这里似乎有点矫枉过正:

#include <iostream>

struct FoodType {
    virtual int getFat() const = 0;
};

struct Apple: FoodType {
    int getFat() const { return 2; }
};

struct Banana: FoodType {
    int getFat() const { return 3; }
};

void eatFood(const FoodType& type) { std::cout << "I ate " << type.getFat() << " grams of fat\n"; }

int main() {
    Apple apple;
    eatFood(apple);
    Banana banana;
    eatFood(banana);
}