How use template function for a templated functor-CodePudding

I need to sort std::pair, the second element are different struct, ideally i would like to have a single elegant function but i could also do overloaded function, unfortunately even with overloaded functions its not compiling because of : <unresolved overloaded function type>

typedef struct{
    int x, y;
} myStruct;

template <typename S> static bool comp(std::pair<int, S> pair1, std::pair<int, S> pair2){
    
    auto val1 = std::get<int>(pair1);
    auto val2 = std::get<int>(pair2);
    
    return val1 < val2;
}


int main(){
    
    myStruct test;
    
    std::vector<std::pair<int, myStruct>> v {{9,test}, {2,test}, {4,test}, {3,test}, {1,test}};
    
    std::sort(v.begin(), v.end(), comp); 

}

CodePudding user response：

The main idea is OK, but there are only one problems with this code.

The sort function MUST know the type of the functor since it cannot be deduced, so you should write:

std::sort(v.begin(), v.end(), comp<myStruct>);

in order to let your compiler know what function to "instantiate" within the Two-phase name lookup

CodePudding user response：

std::sort(v.begin(), v.end(), comp);

comp is a name of a template. comp is not a function, whose name evaluates to a function pointer. comp is a template. The third parameter to std::sort is not a template name. It's a callable object, of some kind. There are certain requirements for that callable object, but that's immaterial here.

How do you make a callable object out of a template? Instantiate this template. How do you instantiate a template? You specify what it's parameters are.

std::sort(v.begin(), v.end(), comp<myStruct>);

In many instances template parameters can be deduced, but not in this case, for rather complicated reasons.