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Consider the following program:

template <typename T>
T foo(const T& x);

template <>
char const * foo<char const *>(char const *& x);

where I'm just using specializing const char*. The compiler won't accept this! It tells me (GodBolt.org):

<source>:5:14: error: template-id 'foo<const char*>' for 'const char* foo(const char*&)' does not match any template declaration
    5 | char const * foo<char const *>(char const *& x)
      |              ^~~~~~~~~~~~~~~~~
<source>:4:3: note: candidate is: 'template<class T> T foo(const T&)'
    2 | T foo(const T& x);
      |   ^~~

Why isn't it accepting my specialization? Am I misunderstanding how template specialization works?

einpoklum
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1 Answers1

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Your understanding of templates is fine, but you're getting confused by C++ type syntax, and by a finer point of reference semantics.

You see, your specializing function needs to take a const T&, for T being char const *. You wrote const char*& - but in this type, it is only the pointed-to data that is const, while the reference is non-const.

Now, you might ask, "What do you mean, the reference is non-const? It's a reference, it can't change; and I can't change anything using the pointer, since it points to const data!" ... well, you're mostly right, but - you can assign to a non-const reference to a pointer, can't you? And that will change the pointer variable's value at the call site.

What you need to write is:

template <>
char const * foo<char const *>(char const * const & x);

i.e. with a rather-redundant extra const indicator, applying to the reference rather than the character data. This will compile (GodBolt.org).

einpoklum
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