--- Comment #9 from roberto dot gordo at gmail dot com 2007-03-17 17:58
---
I would like to apologize for my faults in gcc bug report 31166, specially to
the people who responded, and fully acknowledge my error. I've misunderstood
your responses. Now, while reading them again,
--- Comment #8 from roberto dot gordo at gmail dot com 2007-03-14 12:29
---
I'm still unable to match the behavior of gcc with the ISO C standard. I will
try to explain myself.
The reason for which gcc produces different results with hex constants is now
clear. Also, in the foll
--- Comment #7 from roberto dot gordo at gmail dot com 2007-03-14 09:40
---
That's OK, it is not a bug, sorry.
--
roberto dot gordo at gmail dot com changed:
What|Removed |
--- Comment #6 from roberto dot gordo at gmail dot com 2007-03-14 09:27
---
I think I've found something. According to the ISO C standard, a decimal
constant without suffixes should ALWAYS be signed int (or signed long long if
it does not fit), but never be unsigned! An oct
--- Comment #5 from roberto dot gordo at gmail dot com 2007-03-14 08:52
---
> unsigned is never promoted, it always stays unsigned.
Sorry to insist, but I'm still not convinced. Please, see these examples,
compiled with -std=c99.
{
unsigned u;
int i;
u = 1; /* th
--- Comment #2 from roberto dot gordo at gmail dot com 2007-03-13 22:27
---
I do not agree at all. Please, read.
So 0x8000 is unsigned because does not fit on an int type. That's OK. If
negating it gives an unsigned int of the same value, then, how do you explain
tha
-
Summary: Integer hex constant does not follow promoting rules
Product: gcc
Version: 4.1.2
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c
AssignedTo: unassigned at gcc dot gnu dot org
Rep
