friend

每日一句

由 你离开我真会死。 提交于 2019-12-30 05:11:22
许多或是优美或是深刻的句子，特开此贴，以作收藏。 2014 Mar 21 Get a livelihood, and then practice virtue. 先谋生而后修身。 Apr 19 因为不安而频频回首 无知地索求 羞耻于求救 不知疲倦地翻越每一个山丘 -- 来自李宗盛的《山丘》 临渊羡鱼，不如退而结网。（台上的光鲜，都是由台下的汗水换来的。） Apr 21 Friends are lost by calling often and calling seldom. 交往过密过疏，都会失去朋友。 Jun 26 Being different is one of the most beautiful things on earth. [与众不同是世界上最美好的事情。] Jun 27 Leopards don't change their spots. [本性难移.] Jun 28 One man, no man. [个人是渺小的。] Jun 30 Accidents will happen. [天有不测风云。] You can say that again. [谁说不是呢。] Jul 3 A bosom friend afar brings a land near. [海内存知己，天涯若比邻。a bosom friend: 知心朋友] Jul 4 A hedge

How to allow template function to have friend(-like) access?

由 谁都会走 提交于 2019-12-28 20:34:59
问题 How does one modify the following code to allow template function ask_runUI() to use s_EOF without making s_EOF public? #include <string> #include <iostream> #include <sstream> #include <vector> class AskBase { protected: std::string m_prompt; std::string m_answer; virtual bool validate(std::string a_response) = 0; public: AskBase(std::string a_prompt):m_prompt(a_prompt){} std::string prompt(){return m_prompt;} std::string answer(){return m_answer;} static int const s_EOF = -99; static int

Friend scope in C++

由 大城市里の小女人 提交于 2019-12-27 17:06:08
问题 If I have three classes, A, B, C. A and B are friends (bidirectionally). Also, B and C are friends (bidirectionally). A has a pointer to B and B has a pointer to C. Why can't A access C's private data through the pointer? Just to clarify: This is a pure theoretical C++ language question, not a design advice question. 回答1: Friendship in C++ is not transitive: John is a friend of mine and he can use my wireless connection any time (I trust him). John's friend Tim though is a waster and though

C++: friend as main in class

由 こ雲淡風輕ζ 提交于 2019-12-25 19:36:25
问题 Can main function become friend function in C++ ? #include "stdafx.h" #include <iostream> using namespace std; class A { public: A():i(10){} private: int i; friend int main(); }; int main() { A obj; cout<<obj.i; return 0; } 回答1: Can main function become friend function in C++ ? Yes, it can. The friend declaration in your class A grants function main() the right of accessing the name of its non-public data members (in this case, i ): friend int main(); The object obj is default-constructed,

C++: friend as main in class

由 杀马特。学长 韩版系。学妹 提交于 2019-12-25 19:35:39
问题 Can main function become friend function in C++ ? #include "stdafx.h" #include <iostream> using namespace std; class A { public: A():i(10){} private: int i; friend int main(); }; int main() { A obj; cout<<obj.i; return 0; } 回答1: Can main function become friend function in C++ ? Yes, it can. The friend declaration in your class A grants function main() the right of accessing the name of its non-public data members (in this case, i ): friend int main(); The object obj is default-constructed,

Easier way to write encapsulated parent/child data structure?

由 拜拜、爱过 提交于 2019-12-25 09:03:36
问题 From time to time I find myself often writing a data structure of "parents" and "children", where: A parent has references to 0 to N distinct children. A child has a reference to 0 parents or 1 parent. The reference must be mutual. For any given parent, any child that it references must also reference the given parent back. For any given child, the parent that it references must reference the given child back. It's impossible to violate the above rules through use of members accessible from

getting facebook friends list who have installed an app

由 时间秒杀一切 提交于 2019-12-25 05:35:40
问题 It's possible to know the facebook friends that they have installed a my fb app? I have tried some ways directly using the online facebook developer tool but I have not been able to find a solution. I have tried both Graph API and FQL 回答1: Using FQL: SELECT uid, name, is_app_user FROM user WHERE uid IN (SELECT uid2 FROM friend WHERE uid1=me()) AND is_app_user=1 来源： https://stackoverflow.com/questions/14480748/getting-facebook-friends-list-who-have-installed-an-app

C++, using a key class as a key of access for a group of classes

由 扶醉桌前 提交于 2019-12-24 10:38:12
问题 I've seen a pattern like this in some project: class AccessKey{ // a group of classes called privilegedClasses friend class foo; friend class bar; // friend class other classes in privilegedClasses... private : AccessKey(){}; /*! private constructor */ static const AccessKey key; /*! private object, only for friend classes */ } This is a class that only privilegedClasses can access to it and have an object of it. Now consider someone is writing a function and want to limit access to that

Error C2679 binary '<<': no operator found which takes a right-hand operand of type 'T' (or there is no acceptable conversion)

由 核能气质少年 提交于 2019-12-24 07:29:32
问题 I'm having problems with my friend function within my template class. For some reason it doesn't like the fact that I'm trying to use a variable that is type T in an operator overloading friend function. #include <iostream> #include <fstream> #include <string> template <typename T> class LL { struct Node { T mData; Node *mNext; Node(); Node(T data); }; private: Node *mHead, *mTail; int mCount; public: LL(); ~LL(); bool insert(T data); bool isExist(T data); bool remove(T data); void

Linker error when operator== is a friend [duplicate]

由 杀马特。学长 韩版系。学妹 提交于 2019-12-24 02:37:29
问题 This question already has answers here : overloading friend operator<< for template class (5 answers) Closed 4 years ago . The following code is a minimum code to reproduce my problem. When I try to compile it, the linker can not find operator== for Config : Undefined symbols for architecture x86_64: "operator==(Config<2> const&, Config<2> const&)", referenced from: _main in test2.o The operator== is a friend of Config . BUT when I do no longer declare operator== as a friend, the code
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