--- Additional Comments From schlie at comcast dot net 2005-01-15 02:38
---
(In reply to comment #5)
Yes, I apologize, I misread the the report/answer, you are of course correct.
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=19437
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-01-14
19:50 ---
(In reply to comment #4)
> Then either the semantic definition of the middle-end's w = (x ? y : z) needs
> to preserve y : z types,
> or it can not be considered equilvelent to if(x){ w = y; w = z;}, it wou
--- Additional Comments From schlie at comcast dot net 2005-01-14 19:40
---
(In reply to comment #3)
> int val = i ? i : -1;
>
> we convert it to unsigned because it gets promoted to unsigned because i is
> unsigned and then
> there is an implicit cast to int. so the type of "
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-01-14
14:30 ---
int val = i ? i : -1;
we convert it to unsigned because it gets promoted to unsigned because i is
unsigned and then there is
an implicit cast to int. so the type of "i ? i : -1" is unsigned and t
--- Additional Comments From oliverst at online dot de 2005-01-14 14:26
---
Ah, OK. Why is there warning at all. "-1" is being assigned to an "int" and not
to an "unsigned int".
Writing it without the "? :" operator, the following code should be the same.
unsigned int i = 0;
int val;
if
--- Additional Comments From pinskia at gcc dot gnu dot org 2005-01-14
13:19 ---
The warning is correct.
To get a warning in C you have to use -Wconversion.
pr19437.c:4: warning: negative integer implicitly converted to unsigned type
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What|Removed
