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class variables is shared across all instances in python? [duplicate]

由 左心房为你撑大大i 提交于 2019-11-26 20:29:52
This question already has an answer here: How to avoid having class data shared among instances? 8 answers I started coding in python a week ago, it is my mistake i started coding using oops,classes and objects that soon. I assumed my C++ proficiency will help.... I got bit by the following code class A: var=0 list=[] def __init__(self): pass Here to my surprise, var and list are kinda global variable, it is shared across all instances it seems.... What I thought was it was different across all the instances..... It took me half a day to figure out that.... It does not make even slightest

How can I declare a member vector of the same class?

由 有些话、适合烂在心里 提交于 2019-11-26 18:25:19
问题 Why on earth does the following piece of code work? struct A { std::vector<A> subAs; }; A is an incomplete type, right? If there was a vector of A*s I would understand. But here I don't understand how it works. It seems to be a recursive definition. 回答1: This paper was adopted into C++17 which allows incomplete types to be used in certain STL containers. Prior to that, it was Undefined Behavior. To quote from the paper: Based on the discussion on the Issaquah meeting, we achieved the

Cycle in the struct layout that doesn't exist

由 五迷三道 提交于 2019-11-26 16:50:54
问题 This is a simplified version of some of my code: public struct info { public float a, b; public info? c; public info(float a, float b, info? c = null) { this.a = a; this.b = b; this.c = c; } } The problem is the error Struct member 'info' causes a cycle in the struct layout. I'm after struct like value type behaviour. I could simulate this using a class and a clone member function, but I don't see why I should need to. How is this error true? Recursion could perhaps cause construction forever

C++ class member function callback

由 霸气de小男生 提交于 2019-11-26 16:48:21
问题 I have the following problem. I have a function from an external library (which cannot be modified) like this: void externalFunction(int n, void udf(double*) ); I would like to pass as the udf function above a function member of an existing class. Please look at the following code: // External function (tipically from an external library) void externalFunction(int n, void udf(double*) ) { // do something } // User Defined Function (UDF) void myUDF(double* a) { // do something } // Class

C++ inline member function in .cpp file

由 雨燕双飞 提交于 2019-11-26 16:24:39
I know that inline member functions by definition should go into the header. But what if it's not possible to put the implementation of the function into the header? Let's take this situation: File A.h #pragma once #include "B.h" class A{ B b; }; File B.h #pragma once class A; //forward declaration class B{ inline A getA(); }; Due to the circular include I have to put the implementation of getA into B.cpp #include "B.h" #include "A.h" inline A B::getA(){ return A(); } Will the compiler inline getA ? If so, which inline keyword is the significant one (the one in the header or the one in the

What kind of prefix do you use for member variables?

由 假如想象 提交于 2019-11-26 13:03:39
问题 No doubt, it\'s essential for understanding code to give member variables a prefix so that they can easily be distinguished from \"normal\" variables. But what kind of prefix do you use? I have been working on projects where we used m_ as prefix, on other projects we used an underscore only (which I personally don\'t like, because an underscore only is not demonstrative enough). On another project we used a long prefix form, that also included the variable type. mul_ for example is the prefix

php static function

由 岁酱吖の 提交于 2019-11-26 12:52:50
问题 I have a question regarding static function in php. let\'s assume that I have a class class test { public function sayHi() { echo \'hi\'; } } if I do test::sayHi(); it works without a problem. class test { public static function sayHi() { echo \'hi\'; } } test::sayHi(); works as well. What are the differences between first class and second class? What is special about a static function? 回答1: In the first class, sayHi() is actually an instance method which you are calling as a static method

Why I can access member functions even after the object was deleted?

由 假装没事ソ 提交于 2019-11-26 11:38:50
问题 I\'m new to C++ and from what I learned so far when you call delete on a pointer that points to something created on the heap then whatever is pointed by that pointer gets erased and the memory is freed, right? However when I tried this on a simple class: class MyClass { int _Id; public: MyClass(int id) : _Id(id) { std::cout << \"$Constructing the damn thing! \" << _Id << std::endl; } ~MyClass() { std::cout << \"?Destructing the damn thing! \" << _Id << std::endl; } void Go_XXX_Your_Self() {

C++ static template member, one instance for each template type?

由 守給你的承諾、 提交于 2019-11-26 11:27:28
问题 Usually static members/objects of one class are the same for each instance of the class having the static member/object. Anyways what about if the static object is part of a template class and also depends on the template argument? For example, like this: template<class T> class A{ public: static myObject<T> obj; } If I would cast one object of A as int and another one as float , I guess there would be two obj , one for each type? If I would create multiple objects of A as type int and

“Incomplete type” in class which has a member of the same type of the class itself

由 |▌冷眼眸甩不掉的悲伤 提交于 2019-11-26 11:22:14
I have a class that should have a private member of the same class, something like: class A { private: A member; } But it tells me that member is an incomplete type. Why? It doesn't tell me incomplete type if I use a pointer, but I'd rather not use a pointer. Any help is appreciated At the time you declare your member, you are still defining the A class, so the type A is still undefined. However, when you write A* , the compiler already knows that A stands for a class name, and so the type "pointer to A" is defined . That's why you can embed a pointer to the type your are defining. The same
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