I am having trouble with displaying opencv Video on a Qlabel. I'm new using opencv and qt, and these week I was trying to make a tiny exercise using a qt button event to display a video capture from opencv to a qlabel of my widget. But weirdly the program says "The program has unexpectedly finished", when I run the code attached below. Please help me cuz for me nothing seems to
我在Qlabel上显示opencv视频时遇到问题。 我是使用opencv和qt的新手，并且这些星期我正在尝试使用qt按钮事件做一个小小的练习来显示从opencv到我的小部件的qlabel的视频捕捉。 但奇怪的是，当我运行下面的代码时，程序显示“程序意外完成”。 请帮助我因为我没有任何错误。 感谢您的时间和来自哥斯达黎加的问候。 PS当我简单地尝试运行没有gui的openCv代码时，我的意思是使用buttonClicked事件和cvShowImage（“Video”，fram
The below works: struct { int v; } vals[] = { {1}, {2} }; Can I do the same thing but instead initialize an std::array ? Edit since so many people are asking 'why' There are some very obvious workarounds (listed in the comments), but I only want to use the type once, so I don't really want it added to my current namespace. I could use a tuple or something similar, but havin
以下工作： struct { int v; } vals[] = { {1}, {2} }; 我可以做同样的事情，而是初始化一个std::array吗？ 因为很多人都在问'为什么' 有一些非常明显的解决方法（在评论中列出），但我只想使用一次类型，所以我不想将它添加到我当前的命名空间。 我可以使用一个元组或类似的东西，但是命名值可以提高清晰度。 如果我构造一个std::array ，我不需要c数组值，所以我不能使用decltype。 我想要做的最干净的
So I tried yesterday to start using std::initializer_list but that wasn't a huge success. There is one of my last try: #include <unordered_map> #include <string> struct XmlState { using U_StateFunc = std::function<void()>; using U_MapStateFunc = std::unordered_map<std::string, U_StateFunc>; U_StateFunc beforeProcess; U_Stat
所以我昨天试着开始使用std :: initializer_list，但那并不是一个巨大的成功。 有我最后的尝试之一： #include <unordered_map> #include <string> struct XmlState { using U_StateFunc = std::function<void()>; using U_MapStateFunc = std::unordered_map<std::string, U_StateFunc>; U_StateFunc beforeProcess; U_StateFunc afterProcess;
I found it quite odd that the following program still compiled fine despite the default constructor being private (4.8.1 g++): class A{ private: A() = default; A(const A&) = default; }; int main(){ A a; } Actually from 8.4.2[2] of the standard (N3242) An explicitly-defaulted function may be declared constexpr only if it would have been implicitly declared as constex
我发现很奇怪，尽管默认的构造函数是private （4.8.1 g ++），但下面的程序仍然可以很好地编译： class A{ private: A() = default; A(const A&) = default; }; int main(){ A a; } 实际上从标准（N3242）的8.4.2 [2] 显式默认函数只有在隐式声明为constexpr的情况下才可以声明为constexpr。 如果它的第一次声明明确违约， - 它应该是公开的， .......... 默认说明符忽略访问规范的目的
With the a function taking std::initializer_list as argument like shown below int sumOf(std::initializer_list<int> numbers) { int sum = 0; for (auto x : numbers) { sum += x; } return sum; } This code works auto sum = sumOf({ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }); but not this int i[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; auto x = sumOf(i); Why does the s
使用std :: initializer_list作为参数的函数，如下所示 int sumOf(std::initializer_list<int> numbers) { int sum = 0; for (auto x : numbers) { sum += x; } return sum; } 此代码有效 auto sum = sumOf({ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }); 但不是这样 int i[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; auto x = sumOf(i); 为什么第二种形式不起作用？ 或者我做错了什么？
In a variant class I'm working on the raw storage is a char array: alignas(/* the strictest alignment of all types of the variant */) char storage[/* ... */]; The assignment operator is something like: template<class X> void operator=(const X &x) { // ...code for clearing the storage and setting the tag for type X... new(storage) X(x); } while the code for getting the store
在一个变体类中，我正在处理原始存储是一个char数组： alignas(/* the strictest alignment of all types of the variant */) char storage[/* ... */]; 赋值运算符如下所示： template<class X> void operator=(const X &x) { // ...code for clearing the storage and setting the tag for type X... new(storage) X(x); } 而获取存储对象的代码是： template<class X> const X &get() { // ..
I'm trying to create a general operator<< for std::ostream and any Iterable type. This is the code: template <class T,template<class> class Iterable> inline std::ostream& operator<<(std::ostream& s,const Iterable<T>& iter){ s << "[ "; bool first=false; for(T& e : iter){ if(first){ first=false; s
我试图创建一个普通的operator<< for std::ostream和任何Iterable类型。 这是代码： template <class T,template<class> class Iterable> inline std::ostream& operator<<(std::ostream& s,const Iterable<T>& iter){ s << "[ "; bool first=false; for(T& e : iter){ if(first){ first=false; s << e; }els
Let's say I have a template that stores an object of type T . I want to pass constructor arguments in order to initialize the data member. Should I use uniform-initialization or direct-initialization with non-curly braces?: template<typename T> struct X { template<typename... Args> X(Args&&... args) : t(std::forward<Args>(args)...) // ? /*
比方说，我有一个存储类型T的对象的模板。 我想传递构造函数参数来初始化数据成员。 我应该使用统一初始化还是使用非花括号直接初始化？： template<typename T> struct X { template<typename... Args> X(Args&&... args) : t(std::forward<Args>(args)...) // ? /* or */ : t{std::forward<Args>(args)...} // ? private: T t; }; 如果我想存储的对象是一个
I'm trying to implement something like "composed icon" in Qt. Target: I need to dynamicaly set color just for the portion of icon. My idea: Compose this icon by two other. One icon that will be coloured as desired (perhaps by ColorizeEffect) and blend it under second icon that acts as overlay layer. Issue: I tried QIconEngine and implementing its paint method. ColorizeEffect
我试图在Qt中实现像“合成图标”之类的东西。 目标：我需要为图标的部分动态设置颜色。 我的想法：另外两个图标组成这个图标。 一个图标将根据需要进行着色（可能通过ColorizeEffect）并将其混合到第二个用作覆盖图层的图标下。 问题：我尝试了QIconEngine并实施其绘图方法。 ColorizeEffect似乎没有工作（即使当我尝试用临时QLabel进行破解 - 当强度设置为0.0时，由它形成的QIcon为空）。 但这不是主要问题。 事情是，
I have a Tree class of the following type class Tree{ private: Node *root; // trie root node unordered_map<int, vector<Node*> > *index; unsigned long long size; int count; public: ................ }; I am having a vector of the following type vector<Tree> vect; With the above declaration I am encountering a
我有以下类型的Tree类 class Tree{ private: Node *root; // trie root node unordered_map<int, vector<Node*> > *index; unsigned long long size; int count; public: ................ }; 我有一个以下类型的向量 vector<Tree> vect; 通过上面的声明，我在遇到一个分段错误时，将许多节点添加到每个树中。 但是，当我使用下