带有 const ref 参数的函数模板特化

Question

The following code compiles fine .

#include <iostream>

struct rgb8{
    uint8_t r() const {return 0;};
};


template<typename L, typename P>
L pixelToLevel(P p) {
    return static_cast<L>(p);
}

template<>
uint8_t pixelToLevel<uint8_t, rgb8>(rgb8 p) {  //       <---------- line X
    return pixelToLevel<uint8_t, uint8_t>(p.r());
}


int main()
{
  pixelToLevel<uint8_t>(rgb8());
  return 0;
}

But if in line XI change rgb8 p to const rgb8& p , it fails to compile.

(The exact compiler error generated depends on whether or not the explicit template argument rgb8 is also changed to const rgb8& .)

How can I get it to compile if I want to pass p by reference rather than by value on line X?

Answer 1

You need to change the template parameter in the declaration of specialization, and template argument when calling it too. Otherwise the primary template would be called instead. eg

template<>
uint8_t pixelToLevel<uint8_t, const rgb8&>(const rgb8& p) {  //       <---------- line X
    return pixelToLevel<uint8_t, uint8_t>(p.r());
}

then

pixelToLevel<uint8_t, const rgb8&>(rgb8());

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---

EDIT

Given pixelToLevel<uint8_t>(rgb8()); , template argument deduction is performed with primary template and P is deduced as rgb8 (it won't be deduced as const rgb8& with current primary template's parameter declaration), then the specialization version won't be called.

You can apply overloading instead of template specialization. eg

template<typename L, typename P>
L pixelToLevel(P p) {
    return static_cast<L>(p);
}

template<typename L>
L pixelToLevel(const rgb8& p) {  //       <---------- line X
    return pixelToLevel<L, uint8_t>(p.r());
}

Then pixelToLevel<uint8_t>(rgb8()); would select the 2nd overload.

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Answer 2

An alternate solution to @songyuanyao's would be

template<typename L, typename P>
L pixelToLevel(const P& p) {
    return static_cast<L>(p);
}