ROI can only be implemented with an rectangle. I however have a contour that I want to set as an ROI. Does anyone have an idea of how I would go about using a contour as an ROI rather than a rectangle? Otherwise if not possible, how could I focus my actions only in pixels in a specific contour? Thanks PS: Sorry for all these OpenCV questions. Just really confused :$ OpenCV supports onl
ROI只能用矩形实现。 然而，我有一个我想设定为ROI的轮廓。 有没有人知道我会如何将轮廓用作ROI而不是矩形？ 否则，如果不可能，我怎么才能将我的动作只集中在特定轮廓的像素中？ 谢谢 PS：对于所有这些OpenCV问题抱歉。 只是非常困惑：$ OpenCV仅支持矩形投资回报率。 但是，要对特定像素进行一些处理，可以使用一些辅助函数。 其中一个是pointPolygonTest（），它告诉你一个给定的像素不属于多边形。 所以你
For example: class Base { virtual void my_function() = 0; }; class Derived : Base { void my_function() override; }; From what I read, the override keyword is used to make sure that we have the correct signature in the function that we are overriding, and it seems to be its only use. However, in the case of a pure virtual function, the compiler would throw an error if we used an incorrect
例如： class Base { virtual void my_function() = 0; }; class Derived : Base { void my_function() override; }; 从我读到的内容来看， override关键字用于确保我们在覆盖的函数中拥有正确的签名，并且它似乎是它的唯一用途。 但是，在纯虚函数的情况下，如果我们在Derived类（或Base类中使用了不正确的签名，取决于人们如何看待事物），编译器会引发错误。 那么，在Derived::my_function()声明的末尾添加override
C++11 added override to ensure that member functions you write that you intend to override base-class virtual functions actually do (or won't compile). But in a large object hierarchy, sometimes you could accidentally end up writing a member function that overrides a base-class virtual when you didn't intend it! For instance: struct A { virtual void foo() { } // because obviously
C ++ 11增加了override以确保您编写的要覆盖基类虚函数的成员函数实际上（或不会编译）。 但是在一个大对象层次结构中，有时你可能会意外地写出一个成员函数，当你不打算这个函数时，它就会覆盖基类虚拟器！ 例如： struct A { virtual void foo() { } // because obviously every class has foo(). }; struct B : A { ... }; class C : B { private: void foo() { // was intended to be a private funct
What is the purpose of the final keyword in C++11 for functions? I understand it prevents function overriding by derived classes, but if this is the case, then isn't it enough to declare as non-virtual your final functions? Is there another thing I'm missing here? 你缺少的东西，正如idljarn在评论中已经提到的那样，如果你从基类中覆盖了一个函数，那么你不可能将它标记为非虚拟的： struct base
C ++ 11中final关键字用于函数的目的是什么？ 我知道它可以防止派生类重写函数，但是如果是这种情况，那么将它声明为非虚拟的final函数还不够吗？ 我在这里错过了另一件事吗？ 你缺少的东西，正如idljarn在评论中已经提到的那样，如果你从基类中覆盖了一个函数，那么你不可能将它标记为非虚拟的： struct base { virtual void f(); }; struct derived : base { void f() final; // virtual as it overrides base::
The C++11 FDIS it says If a virtual function is marked with the virt-specifier override and does not override a member function of a base class, the program is ill-formed. [ Example: struct B { virtual void f(int); }; struct D : B { void f(long) override; // error: wrong signature overriding B::f void f(int) override; // OK }; What if B::f would not have been marked virtual? Is
它说的C ++ 11 FDIS 如果一个虚函数用virt-specifier覆盖来标记，并且不覆盖基类的成员函数，那么这个程序是不合格的。 [例如： struct B { virtual void f(int); }; struct D : B { void f(long) override; // error: wrong signature overriding B::f void f(int) override; // OK }; 如果B::f不会被标记为虚拟？ 那么该计划是否构成不良？ 或者被override然后被忽略。 我在std文本中找不到任何这种情况的
I am investigating performance hotspots in an application which spends 50% of its time in memmove(3). The application inserts millions of 4-byte integers into sorted arrays, and uses memmove to shift the data "to the right" in order to make space for the inserted value. My expectation was that copying memory is extremely fast, and I was surprised that so much time is spent in memmove
我正在调查一个应用程序中的性能热点，该应用程序将其时间的50％用于memmove（3）。 应用程序将数百万个4字节整数插入到已排序的数组中，并使用memmove将数据“向右移动”，以便为插入值腾出空间。 我的期望是复制记忆的速度非常快，我惊讶于花费了太多的时间。 但是后来我认为memmove很慢，因为它正在移动重叠的区域，这必须在紧密的循环中执行，而不是复制大页面的内存。 我写了一个小型的微型基准来发现memcpy和memmove之
Is there any way to memmove a buffer in CUDA in a bitwise manner? Eg, for a buffer with two bytes and a pointer buf -> 00000000 11111111 I would like to shift bit portions left or right given their bit offset. Something like void memmove(void* buf, int from, int bits, int delta) For the buffer above I would expect then: 00000111 11111111 after calling memmove(buf,8,3,-3) I could not
有没有办法以逐位的方式移除CUDA中的缓冲区？ 例如，对于具有两个字节和一个指针的缓冲区 buf -> 00000000 11111111 考虑到位偏移，我想将位部分左移或右移。 就像是 void memmove(void* buf, int from, int bits, int delta) 对于上面的缓冲区，我期望的是： 00000111 11111111 打电话后 memmove(buf,8,3,-3) 我无法在CUDA的文档或Google中找到适当的功能。 请注意：我对cudaMemcpy或普通memmove不感兴趣。 一
Possible Duplicate: Efficient bitwise operations for counting bits or find the right|left most ones I wrote a search algorithm. In order to optimize it, I want to use a 32bit int to mark if number 0~31 could be used. That is, if I have a state State , I could easily fetch all possible number by using: x = State & -State State -= x But actually, I need to know where is the 1 in x (noti
可能重复： 有效的按位操作用于计数位或找到最右边的位 我写了一个搜索算法。 为了优化它，我想使用一个32bit int来标记是否可以使用数字32bit int 。 也就是说，如果我有一个州的State ，我可以通过使用以下方式轻松获取所有可能的号码： x = State & -State State -= x 但实际上，我需要知道x的1是什么（注意x的二进制形式只有1）？ 例如，如果x = 0000 0100 ，我想知道这是第三个。 我知道我可以通过使用for循
Given an application-wide collection of objects, and many unrelated classes that need frequent access to these objects, what is the best way to provide said access? Example: // Object A, stored in collections, used to do useful things class A { ... public: QString property(const QString& propertyName) {return m_properties.value(propertyName);} protected: QHash<QString,QString>
给定一个应用程序范围的对象集合，以及需要频繁访问这些对象的许多不相关的类，提供上述访问的最佳方式是什么？ 例： // Object A, stored in collections, used to do useful things class A { ... public: QString property(const QString& propertyName) {return m_properties.value(propertyName);} protected: QHash<QString,QString> m_properties; } // Collection class, contains methods to: // -
std::tie provides a convenient way to unpack the contents of a tuple in C++ into separately defined variables, like the sample below illustrates int a, b, c, d, e, f; auto tup1 = std::make_tuple(1, 2, 3); std::tie(a, b, c) = tup1; However, if we have a nested tuple like below auto tup2 = std::make_tuple(1, 2, 3, std::make_tuple(4, 5, 6)); attempting to compile the code std::tie(a, b, c, std
std::tie提供了一种方便的方式来将C ++中元组的内容解压到单独定义的变量中，如下面的示例所示 int a, b, c, d, e, f; auto tup1 = std::make_tuple(1, 2, 3); std::tie(a, b, c) = tup1; 但是，如果我们有像下面这样的嵌套元组 auto tup2 = std::make_tuple(1, 2, 3, std::make_tuple(4, 5, 6)); 试图编译代码 std::tie(a, b, c, std::tie(d, e, f)) = tup2; 失败并报错 /tmp/tuple.cpp:10: error: invalid initializati