获取OpenCV图像类型的枚举名称（例如CV_32FC1）？ [英] Getting enum names (e.g. CV_32FC1) of OpenCV image types?

问题描述

在OpenCV的C ++界面中，看起来很容易检查图像的类型。如果你有一个图像 cv :: Mat img = cv :: imread（someImage.xyz），你只需要执行 int theType = img。类型（）。

In the C++ interface to OpenCV, it seems easy enough to check the type of an image. If you have an image cv::Mat img = cv::imread("someImage.xyz"), you just do int theType = img.type().

但是，正如你所料，调用 img.type（）只是给出一个整数，而不是一个枚举名称（例如 CV_32FC1 ）。

However, as you would expect, calling img.type() just gives an integer, an not an enum name (e.g. CV_32FC1).

有没有简单的方法打印出枚举名称（例如 CV_32FC1 ）如果我知道OpenCV枚举的整数值？

Is there an easy way to print out the enum name (e.g. CV_32FC1) if I know the integer value of the OpenCV enum?

推荐答案

据我所知，OpenCV中不存在这样的功能。

To my knowledge, such a function doesn't exist in OpenCV.

我认为你会更好地编写自己的功能来获得这些功能。很多交换机的案例，但我猜这是做的工作。可以找到枚举这里。

I think you would be better off writing your own function to get those. A lot of switch cases but I guess it does the job. The enumeration can be found here.

编辑：

这是你可以用来提取类型。我猜想可能会有一个更有效的方法，但是现在我不能包围我的头。

This is something you could use to extract the types. I am guessing there could be a more efficient method, but I can't wrap my head around it right now.

std::string getImageType(int number)
{
    // find type
    int imgTypeInt = number%8;
    std::string imgTypeString;

    switch (imgTypeInt)
    {
        case 0:
            imgTypeString = "8U";
            break;
        case 1:
            imgTypeString = "8S";
            break;
        case 2:
            imgTypeString = "16U";
            break;
        case 3:
            imgTypeString = "16S";
            break;
        case 4:
            imgTypeString = "32S";
            break;
        case 5:
            imgTypeString = "32F";
            break;
        case 6:
            imgTypeString = "64F";
            break;
        default:
            break;
    }

    // find channel
    int channel = (number/8) + 1;

    std::stringstream type;
    type<<"CV_"<<imgTypeString<<"C"<<channel;

    return type.str();
}

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