[Bug c++/94767] (unsigned bitfield) + (int) operation results in int, not unsigned int.

["jh718.park at samsung dot com" <gcc-bugzilla@gcc.gnu.org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=94767

jh718.park at samsung dot com changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|WAITING                     |RESOLVED
         Resolution|---                         |INVALID

--- Comment #4 from jh718.park at samsung dot com ---
(In reply to Jonathan Wakely from comment #3)
> (In reply to jh718.park from comment #0)
> > For these variables below,
> > 
> > unsigned m_schemeEnd : 26;
> > unsigned m_userStart;
> > 
> > m_userStart == m_schemeEnd + 1
> > 
> > this comparison emits a compiler warning as below.
> > 
> > warning: comparison of integer expressions of different signedness:
> > ‘unsigned int’ and ‘int’ [-Wsign-compare]
> 
> Why do you think that's wrong?
> 
> In [conv.prom] p5 the standard says:
> 
> "A prvalue for an integral bit-field (11.4.9) can be converted to a prvalue
> of type int if int can represent all the values of the bit-field;"
> 
> Since int is wider than 26 bits it can represent all the values of
> m_schemeEnd, so the operands of 'm_schemeEnd + 1' are both promoted to int,
> and the result is an int.

Thank you for your comment, Jonathan.
I understand your point.

I thought that unsigned bitfield should be converted to unsigned
during usual arithmetic conversions without knowing the item,
http://eel.is/c++draft/conv.prom#5.

Then, I will mark this issue as resolved/invalid, and update the bug
https://bugs.webkit.org/show_bug.cgi?id=211044
with the comment you left here.

Thank you for your help:)