How to explicitly declare a cv::Mat type as part of a OpenCV library function

Question

I am currently trying to concatenate cv::Mat objects using cv::hconcat(std::vector<cv::Mat> vector_of_mats, cv::Mat dest). However, I need the destination cv::Mat to be of type CV_32FC1 rather than the default CV_8UC1.

I tried constructing the destination cv::Mat as:

 descriptor = cv::Mat::zeros(tmp.size(), 256, CV_32FC1); cv::hconcat(vec, descriptor);

But that still produces a CV_8UC1 cv::Mat for descriptor. The type of cv::Mat in the vector is CV_8UC1. Is that where my problem lies?

Thanks in advance!

Nicholas Betsworth · Accepted Answer · 2015-04-20 15:42:26Z

Your suspicions are correct! The type of the cv::Mat is taken from the source.

A quick look at the source code of cv::hconcat confirms this:

void cv::hconcat(const Mat* src, size_t nsrc, OutputArray _dst) { .... _dst.create( src[0].rows, totalCols, src[0].type()); // Type taken from src .... } void cv::hconcat(InputArray _src, OutputArray dst) { std::vector<Mat> src; _src.getMatVector(src); hconcat(!src.empty() ? &src[0] : 0, src.size(), dst); }

Is it possible to declare a cv::Mat with a type without specifying a size? I looked at the long list of constructors and couldn't find one, but I thought perhaps there may be some way.
@marcman Absolutely, I assume you know the number of columns (256), so you can simply do descriptors = cv::Mat(0, 256, CV_32FC1), and then call descriptors.push_back(vec), assuming vec also has 256 columns. Note: I'm not sure what sort of trouble you could run in to here if they are different types, never tried it.

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