std::endl : wrong completion

[killerbot]

For the first time (I think), CB::CC allowed me to complete std::endl. But it putted :

Code Select Expand

std::endl()

The "()" shouldn't be there.

[blueshake]

it is something related to auto assert "()",can you turn off this,and try again.if it work ,there maybe is something wrong with the parse for token "endl".

[killerbot]

note that this does not happen during the completion of std::cout !

[ollydbg]

note that this does not happen during the completion of std::cout !

simple answer:
cout is regarded as a variable, and endl is regarded as a function( several token named "endl" exist under std namespace, but all of them were functions). see the image below:





This is due some parser problem... let me check it if I have time.

[Loaden]

Let's see iostream file (LINE: 536)

// [27.6.2.7] standard basic_ostream manipulators
 /**
  *  @brief  Write a newline and flush the stream.
  *
  *  This manipulator is often mistakenly used when a simple newline is
  *  desired, leading to poor buffering performance.  See
  *  http://gcc.gnu.org/onlinedocs/libstdc++/27_io/howto.html#2 for more
  *  on this subject.
 */
 template<typename _CharT, typename _Traits>
   inline basic_ostream<_CharT, _Traits>&
   endl(basic_ostream<_CharT, _Traits>& __os)
   { return flush(__os.put(__os.widen('\n'))); }

This is a function!

But in:ostream.tcc (LINE: 361 or 384)

#if _GLIBCXX_EXTERN_TEMPLATE
 extern template class basic_ostream<char>;
 extern template ostream& endl(ostream&);
 extern template ostream& ends(ostream&);

Here is a variable.

[ollydbg]

But in:ostream.tcc (LINE: 361 or 384)

#if _GLIBCXX_EXTERN_TEMPLATE
 extern template class basic_ostream<char>;
 extern template ostream& endl(ostream&);
 extern template ostream& ends(ostream&);

Here is a variable.

our parser just recognize this "endl" as a function, because it has some grammar mode:

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AAAA BBBB();


[Loaden]

But in:ostream.tcc (LINE: 361 or 384)

#if _GLIBCXX_EXTERN_TEMPLATE
 extern template class basic_ostream<char>;
 extern template ostream& endl(ostream&);
 extern template ostream& ends(ostream&);

Here is a variable.

our parser just recognize this "endl" as a function, because it has some grammar mode:

Code Select Expand


AAAA BBBB();


Looks can only judge of the special？

Index: src/plugins/codecompletion/codecompletion.cpp
===================================================================
--- src/plugins/codecompletion/codecompletion.cpp   (revision 6543)
+++ src/plugins/codecompletion/codecompletion.cpp   (working copy)
@@ -790,7 +790,8 @@
                 items.Add(tmp);
                 if (autoAddParentheses && token->m_TokenKind == tkFunction)
                 {
-                    m_SearchItem[token->m_Name] = token->m_Args.size() - 2;
+                    if (token->m_Name != _T("endl"))
+                        m_SearchItem[token->m_Name] = token->m_Args.size() - 2;
                 }
                 if (token->m_TokenKind == tkNamespace && token->m_Aliases.size())
                 {

[attachment deleted by admin]

[eranif]

std::endl *is* a function.

When people are typing std::endl they are actually passing the pointer of the function to the 'operator <<' of class basic_ostream

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__ostream_type&
      operator<<(__ostream_type& (*__pf)(__ostream_type&))

The same is true for all other similar methods, like std::hex, std::oct, std::dec etc
So if you are going to special handle the 'endl' case, make sure you handle the other methods as well (hex, dec, oct etc.)

Eran

[ollydbg]

std::endl *is* a function.

When people are typing std::endl they are actually passing the pointer of the function to the 'operator <<' of class basic_ostream

Code Select Expand

__ostream_type&
      operator<<(__ostream_type& (*__pf)(__ostream_type&))

The same is true for all other similar methods, like std::hex, std::oct, std::dec etc
So if you are going to special handle the 'endl' case, make sure you handle the other methods as well (hex, dec, oct etc.)

Eran

Thanks eran for your help.
So, my personal question is: how does Codelite deal with this kind of problem??
thanks.

[eranif]

So, my personal question is: how does Codelite deal with this kind of problem??

I am not

The same as codeblocks:
when user types std::endl -> codelite adds the () and shows the function tip. I never find it a problem since in codelite if I delete the open brace, the closing brace is also deleted

Eran

[ollydbg]

So, my personal question is: how does Codelite deal with this kind of problem??

I am not

OK, Now we have two ways:
One is like codelite does

The same as codeblocks:
when user types std::endl -> codelite adds the () and shows the function tip. I never find it a problem since in codelite if I delete the open brace, the closing brace is also deleted

The other is special handling like: (endl, hex, dec, oct etc)

thanks.

[ollydbg]

maybe, a better way is :

checking the "<<" can accept a function pointer, then we can avoid adding "()" in this case. But it is too complex 

[MortenMacFly]

Looks can only judge of the special？

No good. Imagine you have a local variable / method with this name than endl has a different meaning and appending the brackets might be desired. In addition: actually this codecompletion should not have any language specific elements at all... ;-)

[Loaden]

Looks can only judge of the special？

No good. Imagine you have a local variable / method with this name than endl has a different meaning and appending the brackets might be desired. In addition: actually this codecompletion should not have any language specific elements at all... ;-)

Well, we still remain the same.

[Loaden]

maybe, a better way is :

checking the "<<" can accept a function pointer, then we can avoid adding "()" in this case. But it is too complex 

I think this is not possible！