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Home/ Questions/Q 8918549
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T05:46:53+00:00 2026-06-15T05:46:53+00:00

Even by explicitly mentioning that the parameter in the function pointer is const, it

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Even by explicitly mentioning that the parameter in the function pointer is const, it doesn’t seem to be able to convert the function to this type:

#include <iostream>

template <typename T>
class Image{};

template <typename TPixel>
static void
FillImage(const Image<TPixel>* const image){}
//FillImage(Image<TPixel>* const image){} // Replacing the above line with this one compiles fine

int main()
{
  typedef Image<float> ImageType;
  ImageType* image = new ImageType;
  void (*autoFunctionPointer)(const decltype(image)) = FillImage;
  autoFunctionPointer(image);
}

Can anyone explain how to make it do that conversion?

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  1. Editorial Team
    Editorial Team
    2026-06-15T05:46:55+00:00Added an answer on June 15, 2026 at 5:46 am

    The const applies to the pointer.

    So, const decltype(image) is equivalent to ImageType* const and not const ImageType*

    If you change image to

    const ImageType* image = new ImageType;
    

    the first version of FillImage() works as expected.

    To get a const ImageType* you can use std::remove_pointer

    #include <type_traits>
    ...
    void (*autoFunctionPointer)(const std::remove_pointer<decltype(image)>::type *) = FillImage;
    
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