[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #1 from jellegeerts at gmail dot com 2010-08-31 20:02 --- Created an attachment (id=21619) -- (http://gcc.gnu.org/bugzilla/attachment.cgi?id=21619action=view) output of `gcc -v -save-temps -std=c99 -O -g -Wall gcctest.c -o gcctest' -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #2 from jellegeerts at gmail dot com 2010-08-31 20:03 --- Created an attachment (id=21620) -- (http://gcc.gnu.org/bugzilla/attachment.cgi?id=21620action=view) output of `gcc -v -save-temps -std=c99 -O -g -Wall gcctest.c -o gcctest' -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #3 from jellegeerts at gmail dot com 2010-08-31 20:03 --- Created an attachment (id=21621) -- (http://gcc.gnu.org/bugzilla/attachment.cgi?id=21621action=view) the `.i' file that GCC created -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #4 from jellegeerts at gmail dot com 2010-08-31 20:04 --- Created an attachment (id=21622) -- (http://gcc.gnu.org/bugzilla/attachment.cgi?id=21622action=view) `.i' file that GCC created -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #5 from jakub at gcc dot gnu dot org 2010-08-31 20:13 --- That's because the whole foo function doesn't have any side-effects, so it is optimized away completely. -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #6 from jellegeerts at gmail dot com 2010-08-31 20:14 --- It also happens in functions that do have side-effects. I can give you an example if you want? -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #7 from jellegeerts at gmail dot com 2010-08-31 20:32 ---
Updated code snippet, GCC doesn't warn here either if we leave `#if 0' as-is,
even though the function foo() may have side-effects.
#include stdio.h
static int array[32];
#if 0 // If '#if 1' is used, GCC warns correctly about the use of uninitialized
variable 'i' below.
int foo(void);
int foo(void)
#else
static int foo(void)
#endif
{
for (int i; i 32; ++i)
{
if (array[i])
return 1;
}
return 0;
}
int main(void)
{
foo();
return 0;
}
--
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #8 from manu at gcc dot gnu dot org 2010-08-31 20:37 --- (In reply to comment #7) Updated code snippet, GCC doesn't warn here either if we leave `#if 0' as-is, even though the function foo() may have side-effects. No, the function below does not have any side-effects. The result of the program is the same whether the function runs or not because the return value is ignored. To have a side-effect you need to write/print, change global memory, or affect the return value of the program. Until then, this is invalid. Please, reopen when you have a valid testcase (there are bugs in Wuninitialized, so I wouldn't be surprised if you find one). -- manu at gcc dot gnu dot org changed: What|Removed |Added CC||manu at gcc dot gnu dot org Status|UNCONFIRMED |RESOLVED Resolution||WORKSFORME http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #9 from jellegeerts at gmail dot com 2010-08-31 20:47 --- Okay. :) Though, why does GCC warn when we have `#if 1', and not if we have `#if 0'? Just curiosity... -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #10 from jellegeerts at gmail dot com 2010-08-31 20:49 --- Also, it seems a bit questionable to not warn when it is clearly(?) not the developers intent to use an uninitialized variable. What is the rationale behind this? Is it a pragmatic thing? -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #11 from jakub at gcc dot gnu dot org 2010-08-31 20:53 --- Because when foo is not static, it has to be compiled. If it is static, GCC figures it is a pure function (only reads memory and arguments and computes from it its return value) and as the result in main of the function isn't used, that call is optimized away very early. As foo isn't used anywhere else, it isn't compiled at all. -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
[Bug c/45467] gcc won't warn about an uninitialized value
--- Comment #12 from jellegeerts at gmail dot com 2010-08-31 20:54 --- Thanks. :) -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45467
