8.9 Explain the data structures and algorithms that you would use to design an in-memory file system. Illustrate with an example in code where possible.
这道题让我们设计一个内存文件系统，咋一听感觉挺吓人啊，像是很底层的东西，但其实只是一道很普通的OOB的题目而已。根据书上所述，在一个简化的文件系统，由文件和目录构成的，而文件类File和目录类Directory都是由一个入口类Entry派生而来的，可参见如下代码：
class Directory;class Entry {public: Entry(string n, Directory *p) { _name = n; _parent = p; _created = getCurrentTimeMillis(); } static long getCurrentTimeMillis() { time_t res = time(NULL); localtime(&res); return (long)res; } bool deleteEntry() { if (_parent == nullptr) return false; return _parent->deleteEntry(this); } virtual int size() = 0; string getFullPath() { if (_parent == nullptr) return _name; else return _parent->getFullPath() + "/" + _name; } long getCreationTime() { return _created; } long getLastUpdatedTime() { return _lastUpdated; } long getLastAccessedTime() { return _lastAccessed; } void changeName(string n) { _name = n; } string getName() { return _name; }protected: Directory *_parent; long _created; long _lastUpdated; long _lastAccessed; string _name;};class File: public Entry {public: File(string n, Directory *p, int sz): Entry(n, p) { _size = sz; } int size() { return _size; } string getContents() { return _content; } void setContents(string c) { _content = c; }private: string _content; int _size;};class Directory: public Entry {public: Directory(string n, Directory *p): Entry(n, p) {} int size() { int size = 0; for (auto a : _contents) { size += a->size(); } return size; } int numberOfFiles() { int cnt = 0; for (auto a : _contents) { if (Directory *d = dynamic_cast (a)) { ++cnt; cnt += d->numberOfFiles(); } else if (File *f = dynamic_cast (a)) { ++cnt; } } return cnt; } bool deleteEntry(Entry *entry) { for (vector ::iterator it = _contents.begin(); it != _contents.end(); ++it) { if (*it == entry) { _contents.erase(it); } } } void addEntry(Entry *entry) { _contents.push_back(entry); }protected: vector _contents; vector getContents() { return _contents; }}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。