Is it possible to define some kind of template that can create a generic comparable operator for structs? For example is it possible for something like this? struct A { int one; int two; int three; }; bool AreEqual() { A a {1,2,3}; A b {1,2,3}; return ComparableStruct<A>(a) == ComparableStruct<A>(b); } All this does is a field by field comparison of the st
是否可以定义某种可以为结构创建通用可比操作符的模板？ 例如，这可能是这样的吗？ struct A { int one; int two; int three; }; bool AreEqual() { A a {1,2,3}; A b {1,2,3}; return ComparableStruct<A>(a) == ComparableStruct<A>(b); } 所有这些都是通过结构的字段比较来进行的。 您可以假定所有字段都是基本类型或者运算符==重载。 我有很多像这样的结构，如果我可以将它放在
If I try to compile: class Outer { class Inner { int t; }; public: Inner inner_; bool operator ==(Outer rightSide); }; bool Outer::operator ==(Outer rightSide) { if (inner_ == rightSide.inner_) return true; return false; } I get an error: /home/martin/Projects/Experimentation/Various-build-desktop/../Various/main.cpp:18: error: no match for ‘oper
如果我尝试编译： class Outer { class Inner { int t; }; public: Inner inner_; bool operator ==(Outer rightSide); }; bool Outer::operator ==(Outer rightSide) { if (inner_ == rightSide.inner_) return true; return false; } 我收到一个错误： /home/martin/Projects/Experimentation/Various-build-desktop/../Various/main.cpp:18: error: no match for ‘operato
I am still new to C++ (programming in general)and forgive me if this question is stupid or has been asked numerously. Here is the question..Let's say there are two objects A and B under the same class. eg class Fruit{ int apple; int banana; fruit(int x, int y){ apple=x; banana=y; } } Fruit A(1,1); Fruit B(1,1); If I want to check if content from Obje
我仍然对C ++不熟悉（一般编程），并且如果这个问题很愚蠢或者被无数次问到，请原谅我。 这里有个问题。比方说，在同一个班上有两个对象A和B. 例如 class Fruit{ int apple; int banana; fruit(int x, int y){ apple=x; banana=y; } } Fruit A(1,1); Fruit B(1,1); 如果我想检查对象A的内容是否与对象B相同，是否必须将每个变量从A比较到B，或者 if(Object A == Object B) ret
I often have classes which provide simple member-by-member comparison: class ApplicationSettings { public: bool operator==(const ApplicationSettings& other) const; bool operator!=(const ApplicationSettings& other) const; private: SkinType m_ApplicationSkin; UpdateCheckInterval m_IntervalForUpdateChecks; bool m_bDockSelectionWidget; // Add future members to operator== }
我经常有类提供简单的逐个成员的比较： class ApplicationSettings { public: bool operator==(const ApplicationSettings& other) const; bool operator!=(const ApplicationSettings& other) const; private: SkinType m_ApplicationSkin; UpdateCheckInterval m_IntervalForUpdateChecks; bool m_bDockSelectionWidget; // Add future members to operator== }; bool ApplicationSettings::opera
I'm using the CPoint class from MFC. There is no explicitly defined assignment operator or copy constructor (AFAIK). Yet, this works: CPoint p1(1, 2), p2; p2 = p1; // p2 now is equal to p1 I'm assuming this is working automagically because of a compiler generated assignment operator. Correct? If so, can I be confident that this isn't doing anything unexpected? In this case CP
我使用MFC中的CPoint类。 没有明确定义的赋值运算符或复制构造函数（AFAIK）。 然而，这是有效的： CPoint p1(1, 2), p2; p2 = p1; // p2 now is equal to p1 我假设这是自动工作的，因为编译器生成的赋值运算符。 正确？ 如果是这样，我能确信这不会出乎意料吗？ 在这种情况下，CPoint非常简单，我认为一切都很好，但总的来说，这让我很担心。 做更好的形式是： p2.SetPoint(p1.x, p2.x); -cr 这是安全的 - 如果
After seeing this question, my first thought was that it'd be trivial to define generic equivalence and relational operators: #include <cstring> template<class T> bool operator==(const T& a, const T& b) { return std::memcmp(&a, &b, sizeof(T)) == 0; } template<class T> bool operator<(const T& a, const T& b) { return std::memcmp(&a,
看到这个问题后，我首先想到的是定义泛型等价关系运算符是微不足道的： #include <cstring> template<class T> bool operator==(const T& a, const T& b) { return std::memcmp(&a, &b, sizeof(T)) == 0; } template<class T> bool operator<(const T& a, const T& b) { return std::memcmp(&a, &b, sizeof(T)) < 0; } using namespace std::rel_ops会
I wanted to know Why compiler provides default copy constructor..Whats the strategy behind that idea. Thanks in Advance. From a related (but not same) question - Why don't C++ compilers define operator== and operator!=?: Stroustrup said this about the default copy constructor in "The Design and Evolution of C++" (Section 11.4.1 - Control of Copying): I personally consider it
我想知道为什么编译器提供了默认的拷贝构造函数。这个想法背后的策略是什么。 提前致谢。 从一个相关的（但不是相同的）问题 - 为什么C ++编译器不定义运算符==和运算符！= ?: Stroustrup在“C ++的设计和发展”（第11.4.1节 - 复制的控制）中对默认的拷贝构造函数进行了说明： 我个人认为不幸的是复制操作是默认定义的，我禁止复制许多课程的对象。 但是，C ++从C继承了它的默认赋值和拷贝构造函数，并且它们经常被使用
I've come across bizarre error related to reinterpret_cast. Just look at below code: int* var; reinterpret_cast<void const **>(&var); error in VSC++2010: error C2440: 'reinterpret_cast' : cannot convert from 'int ** ' to 'const void ** ' error in gcc 4.1.2: reinterpret_cast from type 'int** ' to type 'const void** ' casts away constness
我遇到了与reinterpret_cast有关的奇怪错误。 请看下面的代码： int* var; reinterpret_cast<void const **>(&var); VSC ++ 2010错误：错误C2440：'reinterpret_cast'：无法从'int **'转换为'const void **' gcc 4.1.2中的错误：从'int **'类型到类型'const void **'的reinterpret_cast转换为constness gcc 4.6.2中的错误：从'int **'类型的reinterpret_cast
In the C++ Without Fear: A Beginner's Guide That Makes You Feel Smart book, and in chapter (8), it mentions the following about reinterpret_cast ....converts from one pointer type (int) to another (char*). Because the cast changes the way the data pointed to is interpreted, it is called reinterpret_cast, as opposed to static_cast.* Can you describe this paragraph here? Especially the re
在没有恐惧的C ++中：初学者指南让你感觉到智能书，在第（8）章中，它提到了关于reinterpret_cast ....从一个指针类型（int）转换为另一个（char *）。 由于演员改变了所指向的数据的解释方式，因此它被称为reinterpret_cast，而不是static_cast。* 你能在这里描述这个段落吗？ 特别是操作命名方式的原因是什么？ 谢谢。 基本上，reinterpret_cast将特定位置的位模式重新解释为不同类型。 请参阅此处的示例：http://
So if your converting from Void* to Type* or from Type* to Void* should you use: void func(void *p) { Params *params = static_cast<Params*>(p); } or void func(void *p) { Params *params = reinterpret_cast<Params*>(p); } To me static_cast seems the more correct but I've seen both used for the same purpose. Also, does the direction of the conversion matter. ie should I
所以如果你使用从Void *到Type *或从Type *到Void *的转换： void func(void *p) { Params *params = static_cast<Params*>(p); } 要么 void func(void *p) { Params *params = reinterpret_cast<Params*>(p); } 对我来说，static_cast似乎更正确，但我已经看到两者用于相同的目的。 另外，转换的方向是否重要。 即我应该仍然使用static_cast： _beginthread(func,0,static_cast<void*>(params)