How to correctly link opencv3 under ROS Indigo (using CMake)?

Am not being able to link opencv3 that I installed myself separately from ros. Am getting this error ../devel/lib/libirTest.so: undefined reference to `cv::ORB::create(int, float, int, int, int, int, int, int, int)' ../devel/lib/libirTest.so: undefined reference to `cv::calcOpticalFlowPyrLK(cv::_InputArray const&, cv::_InputArray const&, cv::_InputArray const&, cv::_InputOutputArray

如何正确链接ROS Indigo下的opencv3(使用CMake)?

我无法链接自己与ros分开安装的opencv3。 我得到这个错误 ../devel/lib/libirTest.so: undefined reference to `cv::ORB::create(int, float, int, int, int, int, int, int, int)' ../devel/lib/libirTest.so: undefined reference to `cv::calcOpticalFlowPyrLK(cv::_InputArray const&, cv::_InputArray const&, cv::_InputArray const&, cv::_InputOutputArray const&, cv::_OutputArray const&, cv::_

Initializing reference to istream

I am trying to write my program so that it can process either StdIn or a file specified on the command line. I'm doing this by trying to initialize a reference to an istream to either refer to cin or an ifstream , using a conditional. (similar techniques are described here and here) But when I try with ifstream , I seem to get an error that the basic_istream move-constructor is declared

初始化对istream的引用

我正在尝试编写我的程序,以便它可以处理StdIn或在命令行中指定的文件。 我通过尝试初始化一个istream的引用来引用cin或ifstream ,使用条件来做到这一点。 (类似的技术在这里和这里描述) 但是,当我尝试使用ifstream ,我似乎得到一个错误,即basic_istream移动构造函数被声明为protected 。 istream& refToCIN ( cin ); // This is OK const istream& refToFile = ifstream(args[1]);

Why does C++ not have reflection?

This is a somewhat bizarre question. My objectives are to understand the language design decision and to identify the possibilities of reflection in C++. Why C++ language committee did not go towards implementing reflection in the language? Is reflection too difficult in a language that does not run on a virtual machine (like java)? If one were to implement reflection for C++, what will be

为什么C ++没有反射?

这是一个有点奇怪的问题。 我的目标是理解语言设计决策,并确定在C ++中反思的可能性。 为什么C ++语言委员会没有去实施语言反思? 在没有运行在虚拟机上的语言(如java)中,反射太困难了? 如果有人为C ++实现反思,那么挑战是什么? 我想反射的用途是众所周知的:编辑器可以更容易编写,程序代码会更小,可以为单元测试生成模拟等等。 但是,如果您也可以评论反射的用途,那将是非常好的。 在C ++中反射有几个问题

Generic WebService (SOAP) client library for C++

I'm looking for a simple C++ WebService Client Library that can be easily linked into my application. Preferably this library: can be used to access any SOAP WebService (so I can pass the URL, the WebService name, the WebService method and all the arguments as arguments to a function call) can be linked statically in a C++ application (so no DLL's) is freeware or available at a lo

用于C ++的通用WebService(SOAP)客户端库

我正在寻找一个简单的C ++ WebService客户端库,可以很容易地链接到我的应用程序中。 最好这个库: 可以用来访问任何SOAP WebService(所以我可以将URL,WebService名称,WebService方法和所有参数作为参数传递给函数调用) 可以在C ++应用程序中静态链接(所以没有DLL) 是免费的或以低成本可用 可以在我的应用程序中免版税使用 可以查询Web服务的WSDL并返回可用的方法名称,方法的参数及其数据类型 在你们任何人

Are my lambda parameters really shadowing my locals?

I'm dealing with some C code that takes some data, and forwards it to the function passed in: void foo(int* data, void (*fun)(int*)){ (*fun)(data); }; The following works without warning: void bar(int* data){}; int main(){ int data=0; foo(&data,bar); } However, if I use a lambda instead: int main(){ int data=0; foo(&data,[](auto data){}); } I get the following warnin

我的lambda参数是否真的影响我的本地人?

我正在处理一些需要一些数据的C代码,并将其转发给传入的函数: void foo(int* data, void (*fun)(int*)){ (*fun)(data); }; 以下工作没有警告: void bar(int* data){}; int main(){ int data=0; foo(&data,bar); } 但是,如果我使用lambda代替: int main(){ int data=0; foo(&data,[](auto data){}); } 我收到以下警告: warning: declaration of ‘data’ shadows a previous local [-Wshadow] f

std::vector preallocation (size n, capacity n + 2)

My use-case is the following: A vector of size n read from a binary file. Among other variants (iostreams, in my case custom code doing decompression), I can do something with semantics like this: vector<myElem> v; v.resize(n); // from my understanding v now has size n and capacity n fread(v.data(), sizeof(myElem), n, myFile); However, later I will have to (repeatedly) add and remove tw

std ::向量预分配(大小为n,容量为n + 2)

我的用例如下:从二进制文件中读取大小为n的矢量。 在其他变体(iostreams,在我的例子中是自定义代码进行解压缩)中,我可以用这样的语义来做一些事情: vector<myElem> v; v.resize(n); // from my understanding v now has size n and capacity n fread(v.data(), sizeof(myElem), n, myFile); 但是,后来我必须(重复地)向这样的向量添加和移除两个元素。 (虽然这听起来很愚蠢,但它可以给列表添加标记值,使得

What happens to the underlying storage upon vector's copy/move assignment?

For std::vector's copy assignment, is reallocation of storage and shrink of capacity allowed when the source's size is smaller than the destination's capacity? Or is it guaranteed that the reallocation/shrink will not happen (ie always respect previous reserve())? On the other side, if the source's size is bigger than the destination's capacity and a reallocation takes plac

向量复制/移动分配时底层存储会发生什么?

对于std :: vector的复制分配,当源的大小小于目标容量时,是否允许重新分配存储空间和容量的缩小? 还是保证重新分配/收缩不会发生(即始终尊重先前的准备金())? 另一方面,如果来源的规模大于目的地的容量并进行重新分配,是否要求重新分配尊重来源的容量(例如,目的地的新容量应不小于来源的容量,或者甚至要求它们是一样的)? 或者重新分配只是根据新的规模完成它的工作而不考虑源的容量? 至于移动分配,我认为

C++ : Reduce reserved space with std::vector::reserve

Possible Duplicate: How to downsize std::vector? According to cppreference, in std::vector::reserve(size_t n), n is "Minimum amount desired as capacity of allocated storage.". My question is : how to avoid any reallocation knowing only a maximum ? As an example, let's say that I have a list of integers, but I don't know the size of this list (for example this list come fr

C ++:使用std :: vector :: reserve减少保留空间

可能重复: 如何缩小std :: vector? 根据cppreference,在std :: vector :: reserve(size_t n)中,n是“分配存储容量所需的最小数量”。 我的问题是:如何避免只知道最大值的重新分配? 作为一个例子,假设我有一个整数列表,但我不知道这个列表的大小(例如这个列表来自读取文件)。 但是我知道这个列表的最大大小是1000.假设列表的实际大小是800。 目前,我使用std :: reserve(1000),然后是一个push_back()循

How many levels of pointers can we have?

How many pointers ( * ) are allowed in a single variable? Let's consider the following example. int a = 10; int *p = &a; Similarly we can have int **q = &p; int ***r = &q; and so on. For example, int ****************zz; The C standard specifies the lower limit: 5.2.4.1 Translation limits 276 The implementation shall be able to translate and execute at least one progr

我们可以有多少级别的指针?

单个变量允许多少个指针( * )? 我们来考虑下面的例子。 int a = 10; int *p = &a; 同样我们可以有 int **q = &p; int ***r = &q; 等等。 例如, int ****************zz; C标准规定了下限: 5.2.4.1翻译限制 276该实施应能翻译并执行至少一个程序,该程序至少包含以下每个限制的一个实例:[...] 279 - 12指针,数组和函数声明符(以任何组合方式)修改声明中的算术,结构,联合或无效类型 上

Why does the VS2008 std::string.erase() move its buffer?

I want to read a file line by line and capture one particular line of input. For maximum performance I could do this in a low level way by reading the entire file in and just iterating over its contents using pointers, but this code is not performance critical so therefore I wish to use a more readable and typesafe std library style implementation. So what I have is this: std::string line;

为什么VS2008 std :: string.erase()移动它的缓冲区?

我想逐行阅读一个文件并捕获一个特定的输入行。 为了获得最大性能,我可以通过读取整个文件并使用指针迭代其内容来实现此目的,但此代码不是性能关键,因此我希望使用更具可读性和类型安全性的std库样式实现。 所以我拥有的是这样的: std::string line; line.reserve(1024); std::ifstream file(filePath); while(file) { std::getline(file, line); if(line.substr(0, 8) == "Whatever") { // Do