c++ - 使用抽象来制作翻译器

Question

给定

class Allocator {
  public:
   virtual char *allocate(unsigned int size)=0;
      // EFF: allocates a character buffer of size characters
    virtual void dispose(char *buf)=0;
      // REQ: buf was allocated by this allocator
      // EFF: release memory previously allocated.
};
class Translator {
  public:
    virtual char *operator()(const char *s, Allocator &a) = 0;
      // EFF: returns a translation of the C-string s as
      //      another C-string, where the translation 
      //      C-string is allocated by a.
};

假设您要实现以下内容:

void printTranslatedArgs(int argc, char *argv[], 
                         Translator &t, Allocator &a);
  // REQ: argc/argv are in the form of program arguments
  // EFF: prints the translated command line.

我很难理解这是如何工作的，因为分配、处置和运算符是纯虚拟的，所以它们各自的类实际上并没有定义这些函数。

Answer 1

引用支持多态性。这意味着使用函数 printTranslatedArgs 的任何人都需要使用实现所有虚函数的 Translator 和 Allocator 的基类来调用它。你不需要理会函数内部的具体类类型，调用它们就好像它们是任何其他成员函数一样，例如:

char *p = a.allocate(5);