CODEWITHSUNDEEP
c++templatesc++03template-argument-deduction
Reverent Lapwing
Reverent Lapwing

Reputation: 337

Confused by template argument deduction

I've created the following struct:

template<class T>
struct not_equals
{
  not_equals(T d):data(d){};
  bool operator()(T const & in)
  {
    return data != in;
  }
  T data;
};

My expectation was that since I need to pass some value of concrete type d to the constructor, template argument T will be deduced from type of d.

However, this does not happens.

not_equals('0'); // fails with 'missing template arguments'
char zero = '0';
not_equals(zero); // same as above
not_equals<char>('0'); // compiles without errors

What is the reason for compiler not recognizing type of template argument?

Upvotes: 2

Views: 70

Answers (1)

bolov
bolov

Reputation: 75668

c++17 will allow class template deduction

Until then, you can create a "make" function:

template <class T> auto make_not_equals(const T& d) -> not_equals<T>
{
  return {d};
}

auto z = make_not_equals('0');

This is the C++03 version of the make function:

template <class T> not_equals<T> make_not_equals(const T& d)
{
    return not_equals<T>(d);
}

Unfortunately when you declare a variable you need to spell the whole type with the template because of missing auto feature, but the make function can still be helpful in a deduced context, for instance for a parameter:

template <class T> void foo(not_equals<T> ne);

void test()
{
   foo(make_not_equals('0'));
}

Upvotes: 5

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