Howto: c++ Function Pointer with default values

Question

I have:

typedef void (*RespExtractor) (const cv::Mat & image, cv::Mat & resp);

virtual void predict_image(const cv::Mat & src,
            cv::Mat & img_detect,cv::Size patch_size,
            RespExtractor );

void create_hough_features(const cv::Mat & image, cv::Mat & resp, FeatureParams & params =   FeatureParams() );

How would i define the RespExtractor to accept a function with default parameters, such i can call:

predict_image(im_in,im_out,create_hough_features);

I tried following, with no succes:

typedef void (*RespExtractor) (const cv::Mat & image, cv::Mat & resp,FeatureParams params, FeatureParams());

`FeatureParams & params = FeatureParams()` will lead to an error. — ipc, Mar 18 '12 at 18:03
Possible duplicate of [Function pointers with default parameters in C++](http://stackoverflow.com/questions/2576411/function-pointers-with-default-parameters-in-c) — Jonathan Mee, Jun 23 '16 at 02:44

score 18 · Accepted Answer · answered Mar 18 '12 at 17:55

18

Function pointers themselves can't have default values. You'll either have to wrap the call via the function pointer in a function that does have default parameters (this could even be a small class that wraps the function pointer and has an operator() with default paremeters), or have different function pointers for the different overloads of your functions.

answered Mar 18 '12 at 17:55

pmdj

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Good suggestion about the class example. – Poul K. Sørensen Mar 18 '12 at 18:05
I created a class wrapper, is it possible to make an operator abstract in the base class. – Poul K. Sørensen Mar 18 '12 at 18:28
1

@s093294: it is, but I would advise against it. I would advise using a regular function if it needs be virtual... the question of course would be, why should it be virtual, since it already dispatches to a function pointer ? – Matthieu M. Mar 18 '12 at 19:29
Instead of function pointers i decided to create a simple class with operator() overload and then i just inhered from this class instead and pass that as a parameter instead of a function pointer. i made a virtual method generate, which the operator() calls. Thanks for the help. – Poul K. Sørensen Mar 18 '12 at 23:28

score 6 · Answer 2 · answered Mar 18 '12 at 17:55

6

Default parameters aren't part of the function signature, so you can't do this directly.

However, you could define a wrapper function for create_hough_features, or just a second overload that only takes two arguments:

void create_hough_features(const cv::Mat & image, cv::Mat & resp, FeatureParams & params) {
    // blah
}

void create_hough_features(const cv::Mat & image, cv::Mat & resp) {
    create_hough_features(image, resp, DefaultParams());
}

answered Mar 18 '12 at 17:55

Oliver Charlesworth

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Thanks. I did typedef void (*RespExtractor) (const cv::Mat & image, cv::Mat & resp,FeatureParams &); and predict_image(im_in,im_out,create_hough_features,FeatureParams & par=DefaultParams() ); – Poul K. Sørensen Mar 18 '12 at 18:04
Vent with the suggestion of @pmjordan – Poul K. Sørensen Mar 18 '12 at 18:29

Howto: c++ Function Pointer with default values

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