Function object as template parameter

Question

template <typename elemType, typename Comp = less<elemType> >
class LessThanPred {
    public:
        LessThanPred(const elemType &val) : _val(val){}
        bool operator()(const elemType &val) const
        { return Comp(val, _val); }
        void val(const elemType &newval) { _val = newval; }
        elemType val() const { return _val; }
private:
    elemType _val;};

That's an example from Essential c++. Comp is obviously a function object class name. Why could I use the Comp(val, _val) directly? Normally I think I should firstly define a function object like this: Comp comp, then invoke comp not Comp.

Comp is a type parameter, not a function object class name. By the way, you can actually define a `Comp comp` object and then invoke `comp`. — user2296177, Nov 04 '15 at 19:47
Did you instantiate/use the function call operator of `LessThanPred`? Since it is a template member it gets checked for correctly only when instantiated. — Dietmar Kühl, Nov 04 '15 at 19:50
I'm voting to close this question as off-topic because The citation is not correct — , Nov 04 '15 at 19:58
@DieterLücking: section 6.7 "Template Parameters as Strategy" of "Essential C++" (1st printing) does contain that exact code. — Dietmar Kühl, Nov 04 '15 at 20:00

score 1 · Accepted Answer · answered Nov 04 '15 at 19:56

The code as is compiles because the template members are only checked for semantic correctness when they are instantiated. The code is syntactically well-formed, though. However, when you try to instantiate the function call operator of LessThanPred<T>, you'll get a compiler error. For example, with the version of clang (3.6.1) I'm using I get

less-than.cpp:8:18: error: no matching constructor for initialization of 'std::less<int>'
        { return Comp(val, _val); }
                 ^    ~~~~~~~~~
less-than.cpp:17:25: note: in instantiation of member function 'LessThanPred<int, std::less<int> >::operator()' requested here
    LessThanPred<int>(2)(1);

when trying to use the function as

LessThanPred<int>(2)(1)

For the default situation, could I change the code ' { return Comp(val, _val); }' into { return Comp()(val, _val); }? — yycfox, Nov 04 '15 at 22:26

Function object as template parameter

1 Answers1