What does __STRICT_ANSI__ mean in C (not C++) mode?
The doc says: "The macro `__STRICT_ANSI__' is predefined when the `-ansi'
option is used." I.e. when the compiler is implementing strict ISO C90.
But it is also defined when the compiler is implementing strict ISO C99.
$ touch empty.c
$ gcc -std=c99 -E -dM empty.c | grep '\(__STRICT\|__STDC_V\)'
#define __STRICT_ANSI__ 1
#define __STDC_VERSION__ 199901L
And -ansi is not implied in this mode: -std=c99 -ansi apparently means
strict ISO C90:
$ gcc -std=c99 -ansi -E -dM empty.c | grep '\(__STRICT\|__STDC_V\)'
#define __STRICT_ANSI__ 1
Possible resolutions:
A) Document that __STRICT_ANSI__ can also mean strict ISO C99.
B) Change "gcc -std=c99" to define __STRICT_C99__ instead of __STRICT_ANSI__.
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Summary: __STRICT_ANSI__ meaning incorrectly documented
Product: gcc
Version: 4.1.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: bruno at clisp dot org
GCC build triplet: x86_64-suse-linux
GCC host triplet: x86_64-suse-linux
GCC target triplet: x86_64-suse-linux
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=29994
