使用cv :: imdecode后未修改cv :: Mat外部数据

Question

edit: In trying to give a straight forward example of the problem it appears I left out what was causing the real issue. I have modified the example to illustrate the problem.

I am trying to use opencv to perform operations on a cv::Mat that is composed of external data.

Consider this example:

unsigned char *extern_data = new unsigned char[1280*720*3];
cv::Mat mat = cv::Mat(1280, 720, CV_8UC3, extern_data); //Create cv::Mat external

//Edit - Added cv::imdecode
mat = cv::imdecode(mat,1);

//In real implementation it would be mat = cv::imdecode(image,'1')
// where image is a cv::Mat of an image stored in a mmap buffer

mat.data[100] = 99;

std::cout << "External array: " << static_cast<int>(extern_data[100]) << std::endl;
std::cout << "cv::Mat array: " << static_cast<int>(mat.data[100]) << std::endl;

The result of this is:

> External array: 0
> cv::Mat array: 100

It is clear this external array is not being modified, therefore new memory is being allocated for the cv::Mat array. From my understanding this was not suppose to happen! This should have caused no copy operation, and mat.data should be a pointer to extern_data[0] .

What am I misunderstanding?

Answer 1

So far the way I have got my program to work is to use std::copy. I am still wondering if there is a way to assign the result of cv::imdecode() directly to the external data.

Currently I am using

unsigned char *extern_data = new unsigned char[1280*720*3];
cv::Mat mat = cv::Mat(1280, 720, CV_8UC3, extern_data); //Create cv::Mat external

mat = cv::imdecode(mat,1);

std::copy(mat.data, mat.data + 1280*720*3, extern_data);

I just wish i could figure out how to assign the result of cv::imdecode() directly to extern_data without the additional std::copy line!