Headers without cpp files and include ordering

There was a highly rated response in a question about header ordering with the following suggestion: Good practice: every .h file should have a .cpp that includes that .h first before anything else. This proves that any .h file can be put first. Even if the header requires no implementation, you make a .cpp that just includes that .h file and nothing else. Personally I've never had a p

没有cpp文件的头文件,包括排序

在一个关于标题排序的问题中有一个评价很高的回应,并提出以下建议: 良好的做法:每个.h文件都应该有一个.cpp文件,其中包含.h文件。 这证明任何.h文件都可以放在第一位。 即使头文件不需要实现,你也可以制作一个.cpp文件,其中只包含.h文件。 就个人而言,我从来没有遇到包含没有相应cpp文件的头文件的问题。 这种最佳做法会阻碍什么样的问题? 头文件应该自己编译。 即。 用于测试制作一个只包含头文件的.cpp文

ptr<T> required to know the full definition of T?

I have some code in a header that looks like this: #include <memory> class Thing; class MyClass { std::unique_ptr< Thing > my_thing; }; If I include this header in a cpp that does not include the Thing type definition, then this does not compile under VS2010-SP1: 1>C:Program Files (x86)Microsoft Visual Studio 10.0VCincludememory(2067): error C2027: use of undefined type &

需要知道T的完整定义的ptr <T>?

我在头文件中有一些代码,如下所示: #include <memory> class Thing; class MyClass { std::unique_ptr< Thing > my_thing; }; 如果我将这个头文件包含在不包含Thing类型定义的cpp中,那么这不会在VS2010-SP1下编译: 1> C: Program Files(x86) Microsoft Visual Studio 10.0 VC include memory(2067):error C2027:使用未定义的类型'Thing' 用std::unique_ptr std::shared_ptr替

compile program and run in terminal

I am using Ubuntu 12.04, and I was wondering, is it possible to automatically run c++ program from terminal? It really sucks when you have to use build in console because sometimes I make infinite loops by accident and have to restart sublime text to work again. I am using Sublime text 3. Sublime Text 3 includes two build systems you might be interested in: C++ and Make. The C++.sublime-buil

编译程序并在终端中运行

我使用Ubuntu 12.04,我想知道,是否可以从终端自动运行c ++程序? 当你不得不在控制台中使用build时,它确实很糟糕,因为有时我偶然会发生无限循环,并且必须重新启动崇高的文本才能再次运行。 我正在使用崇高的文字3。 Sublime Text 3包含两个您可能感兴趣的构建系统:C ++和Make。 C++.sublime-build文件如下所示: { "shell_cmd": "g++ "${file}" -o "${file_path}/${file_base_name}"", "file_regex": "^(..[^

Compiling amd64 binary on i386 root with amd64 kernel (Debian)

I have a rather old Debian testing system that has all packages installed as i386. Usually I'm running a PAE kernel (linux-image-3.16.0-4-686-pae:i386). I'm trying to compile a simple C++ program that needs more than 4 GB of memory. I've installed the linux-image-3.16.0-4-amd64:amd64 kernel because I think it is not possible to get more than 3GB of memory on a PAE machine. Unfor

使用amd64内核(Debian)在i386 root上编译amd64二进制文件

我有一个相当老的Debian测试系统,它的所有软件包都安装为i386。 通常我运行的是PAE内核(linux-image-3.16.0-4-686-pae:i386)。 我试图编译一个简单的C ++程序,需要超过4 GB的内存。 我已经安装了linux-image-3.16.0-4-amd64:amd64内核,因为我认为在PAE机器上不可能获得超过3GB的内存。 不幸的是,整个工具链/库仍然是i386。 我想我需要GCC(multilib?)和某些库的amd64版本的特殊味道。 我已经找到了关于如何在

Is there a way to avoid the strict aliasing warning?

I'm using a library (ENet), which uses callbacks. In those callback functions, it passes a struct which contains a void* for user data, for your own use. I'd like to use that variable, but not as a pointer. I don't want to allocate memory just so I can point to it, I'd rather use the space of the void* directly to store a size_t. But, as expected, when I cast the void* variab

有没有办法避免严格的锯齿警告?

我正在使用库(ENet),它使用回调。 在这些回调函数中,它传递一个包含用户数据的void *的结构,供您自己使用。 我想使用该变量,但不是作为指针。 我不想分配内存,所以我可以指出它,我宁愿直接使用void *的空间来存储size_t。 但是,正如所料,当我将void *变量转换为size_t变量时,我得到了严格的别名警告。 回调的结构不提供联合来访问它作为void *之外的其他东西。 我知道我可以完全禁用这个警告,但我宁愿为这个

restrict qualifier on member functions (restrict this pointer)

Note: To clarify, the question is not about the use of the restrict keyword in general, but specifically about applying it to member functions as described here. gcc allows you to use the __restrict__ (the GNU++ equivalent to C99's restrict ) qualifier on a member function, effectively making this a restrict qualified pointer within the function's scope. Where is the beef? Most membe

限制成员函数的限定符(限制这个指针)

注意:为了澄清,这个问题不是关于通常使用restrict关键字,而是关于将其应用于成员函数的具体描述。 GCC允许您使用的__restrict__ (GNU的++相当于C99的restrict )限定符上的成员函数,有效地使this一限制的功能的范围内合格指针。 牛肉在哪里? 大多数成员函数上的其他成员的工作,通过访问它们this ,这是一个T* const (和通常非混淆)。 对于this可能是别名,需要在成员函数中使用第二个指向类型的指针,并且它必须

Does restrict help in C if a pointer is already marked const?

Just wondering: When I add restrict to a pointer, I tell the compiler that the pointer is not an alias for another pointer. Let's assume I have a function like: // Constructed example void foo (float* result, const float* a, const float* b, const size_t size) { for (size_t i = 0; i < size; ++i) { result [i] = a [0] * b [i]; } } If the compiler has to assume that

如果指针已经标记为const,是否会在C中限制帮助?

只是想知道:当我将限制添加到指针时,我告诉编译器该指针不是另一个指针的别名。 假设我有一个函数,如: // Constructed example void foo (float* result, const float* a, const float* b, const size_t size) { for (size_t i = 0; i < size; ++i) { result [i] = a [0] * b [i]; } } 如果编译器必须假定result可能与a重叠,则必须每次重新读取一次。 但是,作为a被标记为const ,编译器还

Histogram Binning of Gradient Vectors

I am working on a project that has a small component requiring the comparison of distributions over image gradients. Assume I have computed the image gradients in the x and y directions using a Sobel filter and have for each pixel a 2-vector. Obviously getting the magnitude and direction is reasonably trivial and is as follows: However, what is not clear to me is how to bin these two componen

梯度矢量的直方图分级

我正在研究一个项目,该项目有一个小组件,要求比较图像渐变上的分布。 假设我已经使用Sobel滤波器在x和y方向上计算了图像梯度,并且为每个像素提供了一个2维向量。 显然,规模和方向是相当无足轻重的,如下所示: 然而,我不清楚的是如何将这两个组件分成二维直方图以获得任意数量的垃圾箱。 我曾考虑过这些方面的内容(用浏览器写的): //Assuming normalised magnitudes. //Histogram dimensions are bins * bins. int

Difference between "Edge Detection" and "Image Contours"

Please have a look at the following code #include <iostream> #include <opencv2/core/core.hpp> #include <opencv2/highgui/highgui.hpp> #include <opencv2/imgproc/imgproc.hpp> using namespace std; using namespace cv; Mat src, grey; int thresh = 10; const char* windowName = "Contours"; void detectContours(int,void*); int main() { src = imread("C:/Users/Public/Pictures

“边缘检测”和“图像轮廓”之间的区别

请看下面的代码 #include <iostream> #include <opencv2/core/core.hpp> #include <opencv2/highgui/highgui.hpp> #include <opencv2/imgproc/imgproc.hpp> using namespace std; using namespace cv; Mat src, grey; int thresh = 10; const char* windowName = "Contours"; void detectContours(int,void*); int main() { src = imread("C:/Users/Public/Pictures/Sample Pictures/Penguins.

pointers to a class in dynamically allocated boost multi

I am pretty new to C++ with Boost. I want an object of class "world" to have an array named "chunk" of type "octreenode". Previously I had an ordinary one-dimensional array, and this worked fine. Now I'm trying to move to using a 3D array with Boost's multi_array functionality, and I'm really not sure what I'm doing wrong. Simplified code: clas

指向动态分配的boost multi中的类

我对Boost的C ++很新颖。 我想让一个“world”类的对象拥有一个名为“chunk”的“octreenode”类型的数组。 以前我有一个普通的一维数组,这工作得很好。 现在我试图转向使用具有Boost的multi_array功能的3D数组,我真的不确定我在做什么错误。 简化代码: class world { public: typedef boost::multi_array<octreenode, 3> planetchunkarray; // a boost_multi for chunks typedef planetchunkarray::index index;