继承类C++

Question

我想像这样调用父构造函数:

Character::Character(int x, int y, int h, int s, int a, char dir) : health(h), strength(s), armor(a), direction(dir){
    if(direction == 'N') {
        Visual('^', x, y);
    } else if(direction == 'S') {
        Visual('v', x, y);
    } else if(direction == 'W') {
        Visual('<', x, y);
    } else if(direction == 'E') {
        Visual('>', x, y);
    }
}

但它不能很好地工作,因为它调用父项的默认构造函数 private

class Visual {
    private:
        Visual();
        Visual(const Visual &);
    protected:
        Position coordinate;
        char chara;
    public:
        Visual(char c, int x, int y);
};

Answer 1

创建一个函数将方向转换为另一个字符,并将其传递给公共构造函数:

namespace { //Anonymous Namespace for functions local to your .cpp files to avoid definition conflicts
    char convert_direction(char c) {
        switch(c) {
        case 'N': return '^';
        case 'S': return 'v';
        case 'E': return '>';
        case 'W': return '<';
        default: return '?';
        }
    }
}

Character::Character(int x, int y, int h, int s, int a, char dir) : 
    Visual(convert_direction(dir), x, y),
    health(h), strength(s), armor(a), direction(dir) 
{
}