c++ grammar question

**bpereira** · 09-29-2008

I am currently validating my c++ parser but i failed to parse c++ header which is cygwin, the file is named "C:\cygwin\lib\gcc\i686-pc-cygwin\3.4.4\include\c++\bits/basic_string.tcc"

in that file there is declaration of assign template function:

Code:

   
template <class _Arg1, class _Arg2, class _Result>
    struct binary_function
    {
      typedef _Arg1 first_argument_type;   ///< the type of the first argument
                                           ///  (no surprises here)
      typedef _Arg2 second_argument_type;  ///< the type of the second argument
      typedef _Result result_type;         ///< type of the return type
    };

  template <class _Tp>
    struct less : public binary_function<_Tp, _Tp, bool>
    {
      bool
      operator()(const _Tp& __x, const _Tp& __y) const
      { return __x < __y; }
    };

  template<typename _CharT, typename _Traits, typename _Alloc>
     basic_string<_CharT, _Traits, _Alloc>&
     basic_string<_CharT, _Traits, _Alloc>::
     assign(const _CharT* __s, size_type __n)
     {
       __glibcxx_requires_string_len(__s, __n);
       if (__n > this->max_size())
	 __throw_length_error(__N("basic_string::assign"));
       if (_M_rep()->_M_is_shared() || less<const _CharT*>()(__s, _M_data())
	   || less<const _CharT*>()(_M_data() + this->size(), __s))
	 return _M_replace_safe(size_type(0), this->size(), __s, __n);
       else
	 {
	   // Work in-place
	   const size_type __pos = __s - _M_data();
	   if (__pos >= __n)
	     traits_type::copy(_M_data(), __s, __n);
	   else if (__pos)
	     traits_type::move(_M_data(), __s, __n);
	   _M_rep()->_M_set_sharable();
	   _M_rep()->_M_length = __n;
	   _M_data()[__n] = _Rep::_S_terminal;  // grr.
	   return *this;
	 }
     }

My problem is :
less<const _CharT*>()(__s, _M_data())

Which i really don't understand, seems like this call explicitly the () operator but I cant find in the c++ iso grammar what are the rule used. Is that code use some gcc extension ?

any help would be greatly appreciated.

**matsp** · 09-29-2008

I don't think it's an extension, it is just calling the operator() with two arguments.

--
Mats

**bpereira** · 09-29-2008

That was my bet too, but official grammar require operator keyword to be used, something like:
less<const _CharT*>operator()(__s, _M_data())

also the name of the class is quite disturbing here, the method is not static thus there should be a object somewhere... or at least use scope operator "::"...

**laserlight** · 09-29-2008

I read that as creating a temporary less<const _CharT*> object, and then invoking operator() of that temporary object with two arguments.

**bpereira** · 09-29-2008

dont you need to use new keyword to create object ?

**laserlight** · 09-29-2008

dont you need to use new keyword to create object ?

Of course not. Perhaps you are getting confused with say, Java?

If you want an example:

Code:

#include <iostream>
#include <string>

class Greet
{
public:
    void operator()(const std::string& name) const
    {
        std::cout << "Hello " << name << "!\n";
    }
};

int main()
{
    // Named Greeter.
    Greet greet;
    greet("bpereira"); // Call operator().

    // Temporary Greeter.
    Greet()("world"); // Call operator().
}

**matsp** · 09-29-2008

Originally Posted by bpereira

dont you need to use new keyword to create object ?

Only if you want to create an object on the heap (aka "Free Store" to use Bjarne's nomenclature).

A temporary object is usually a stack-located object, so it doesn't need "new" to create it.

--
Mats

**bpereira** · 09-29-2008

ok thanks for quick answerer, that fixed my issue, sorry I don't practice c++ often enough.

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