简单的矩阵传递即。 cv :: Mat到OpenCV2.4中的函数 [英] Simple passing of Matrices ie. cv::Mat to functions in OpenCV2.4

问题描述

我是opencv的新手，我对编码中最简单的操作之一有疑问：将值传递给函数。

I am new to opencv and I have a doubt with one of the simplest operations in coding: Passing values to functions.

继承人我想做的事，

1. 在主函数中初始化一个 cv：Mat ，例如，从0到0的值50。


2. 将此矩阵作为参数传递给函数 foo（），然后简单地打印出每个元素的值。矩阵。多数民众赞成吧。

1. Initialize a cv:Mat in the main function with say, values from 0 to 50.
2. Pass this matrix as an argument to a function foo() which in turn simply prints out the values of each element in the matrix. Thats it.

现在我已经在多个地方读过，当这些矩阵通过时，它只是传递的标题，而不是实际数据。那我怎么能够在被调用函数中读出这个矩阵的值，好像整个矩阵都被传递了一样？

Now I have read in multiple places that when these matrices are passed, it is only the header that gets passed, not the actual data. Then how am I still able to read out the values of this matrix in the called function as if the entire matrix was passed?

推荐答案

要了解发生了什么，你需要看看 C ++构造函数 ，更具体地说是 复制构造函数 。

To understand what's going on you need to take a look at C++ constructors, more specifically the copy constructor.

创建 cv时： ：Mat 来自另一个 cv :: Mat ，如：

cv::Mat a = cv::imread("huge.png", 1);
cv::Mat b = a;

复制构造函数（这是一个函数）<$ c $调用c> cv :: Mat 来执行对象的副本。在我谈论复制过程中发生的事情之前，你必须意识到，由于 cv :: Mat 用于存储图像，较大的图像可能占用内存中的数百兆字节。那么 cv :: Mat 的复制构造函数在上面的示例中是复制整个标题（高度，宽度，深度信息等） a 进入 b ，而不是复制的整个数据/像素（可能是几百MB），它只是指向（使用指针） a 的原始数据。

the copy constructor (which is a function) of cv::Mat is called to perform the copy of the object. Before I talk about what happens in the copy procedure, you must realize that as cv::Mat is used to store images, larger images might occupy hundreds of megabytes in memory. So what the copy constructor of cv::Mat does in the example above is copy the entire header (height, width, depth info and others) of a into b, but instead of copying the entire data/pixels of a (which might be hundreds of MB), it just points (using a pointer) to the original data of a.

换句话说，不复制整个图像数据是优化/绩效决策。

现在，请考虑调用函数的代码，并将 cv :: Mat 作为参数传递：

Now, consider this code that calls a function and passes cv::Mat as a parameter:

void do_something(cv::Mat src)
{
    // changing src pixels will be the same as changing orig pixels
}

int main()
{
    cv::Mat orig = cv::imread("huge.png", 1);
    do_something(orig);
    return 0;
}

如果你已经研究过如何将参数传递给函数，你知道调用 do_something（a）; 将 按值传递参数 。这意味着它会尝试将 orig 的内容复制到 src 中，但是，此过程会激活复制构造函数 cv :: Mat 这不会像我刚才解释的那样制作数据的硬拷贝。

If you have studied how to pass parameters to functions, you know that calling do_something(a); will pass the parameter by value. This means that it will try to copy the content of orig into src, however, this procedure activates the copy constructor of cv::Mat which will not make a hard copy of the data as I've just explained.

解决方案这个问题？如果您正在编写 do_something（）而您只想制作 orig 的真实副本，只需创建一个新 cv :: Mat 并调用方法 copyTo（）：

Solution to this problem? If you are writing do_something() and you just want to make a real copy of orig, just create a new cv::Mat and call the method copyTo():

void do_something(cv::Mat src)
{
    cv::Mat real_copy = src.copyTo();
    // operating on the data of real_copy will not affect orig
}

但请记住，如果 src 是~100MB调用 copyTo（）来使真正的副本将占用另一个~100MB的内存，这意味着在单个函数调用中，您的程序从100MB变为200MB。在设计系统时请记住这一点。祝你好运。

But remember, if src is ~100MB calling copyTo() to make the real copy will occupy another ~100MB of memory, which means that in a single function call your program just went from 100MB to 200MB. Keep this in mind when designing your system. Good luck.

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