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C++ - Function template specialization not being called

I have the following code:

template<typename T>
bool validate(const T& minimum, const T& maximum, const T& testValue)
{
    return testValue >= minimum && testValue <= maximum;
}

template<>
bool validate<const char&>(const char& minimum, const char& maximum, const char& testValue)
{
    // Allows comparisons with char arguments, ignoring case    
    // Localize by calling previously defined function
    return validate(toupper(minimum), toupper(maximum), toupper(testValue));
}

The first template is used for any inputted types, and the specialization is for literal characters. The code compiles and runs with a main.cpp to test it, but after testing, I found that the specialization isn't being called. It calls the main template. I can't figure out why.

like image 489
Matt Welke Avatar asked Jan 08 '23 18:01

Matt Welke


1 Answers

The template <> bool validate<const char&> specialization is picked by a compiler when type template parameter T from the primary template is deduced or specified explicitly to be const char&. For a call validate('a', 'b', 'c'), T is deduced to be char, and this doesn't match what the specialization expects.

Either provide a specialization for char (that is, not const char&):

template <>
bool validate<char>(const char& minimum, const char& maximum, const char& testValue)
{
    return validate(toupper(minimum), toupper(maximum), toupper(testValue));
}

or define the overload as a non-template:

bool validate(char minimum, char maximum, char testValue)
{
    return validate(toupper(minimum), toupper(maximum), toupper(testValue));
}
like image 54
Piotr Skotnicki Avatar answered Jan 16 '23 04:01

Piotr Skotnicki