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Specialization of 'template struct std::less' in different namespace

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予麋鹿
予麋鹿 2020-12-03 13:42

I am specializing the \'less\' (predicate) for a data type.

The code looks like this:

template<>
struct std::less
{
   bool
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  • 旧巷少年郎
    旧巷少年郎 (楼主)
    2020-12-03 13:49

    Even though the question has been answered by others with answers on how to specialize std::less (by wrap them in a namespace block) and the right way to do it ( to overload operator <).

    However, C++ now allows (in C++11) to speciliaze the way you did in your first example.

    An explicit specialization shall be declared in a namespace enclosing the specialized template. An explicit specialization whose declarator-id is not qualified shall be declared in the nearest enclosing namespace of the template, or, if the namespace is inline (7.3.1), any namespace from its enclosing namespace set. Such a declaration may also be a definition. If the declaration is not a definition, the specialization may be defined later (7.3.1.2).

    I tried the following code with g++ (8.3.0) on my Ubuntu machine. 

    #include 

#include 
#include 
#include 


class myType {

public:
    myType(int in): i_(in) { }

    int i_;
};

template <>
struct std::less  {
    bool operator()(const myType& a, const myType& b) const
    {
        return a.i_ < b.i_;
    }
};


int main(int argc, char *argv[])
{
    std::map vector = { { 1, "1"}, { 2, "2"}, { 3, "3"}, { 0, "0" } };

    for (auto& i: vector)
        std::cout << i.first.i_ << std::endl;
    return 0;
}

    The above code was compiled with 

    g++ --std=c++11 compare.cpp -Wall

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