The declaration of a function without prototype is currently
silently accepted by sparse but a warning is issued for
'old-style' declarations:
... warning: non-ANSI function declaration ...
However, the difference between these two cases is made by
checking if a ';' directly follow the parentheses. So:
int foo();
is silently accepted, while a warning is issued for:
int foo(a) int a;
but also for:
int foo(), bar();
This last case, while unusual, is not less ANSI than a simple
'int foo();'. It's just detected so because there is no ';'
directly after the first '()'.
Fix this by also using ',' to detect the end of function
declarations and their ANSIness.
Signed-off-by: Luc Van Oostenryck <luc.vanoostenryck@xxxxxxxxx>
---
parse.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/parse.c b/parse.c
index ec145601d..eaa3ac2e2 100644
--- a/parse.c
+++ b/parse.c
@@ -1750,7 +1750,7 @@ static enum kind which_func(struct token *token,
if (next->special == ')') {
/* don't complain about those */
- if (!n || match_op(next->next, ';'))
+ if (!n || match_op(next->next, ';') || match_op(next->next, ','))
return Empty;
warning(next->pos,
"non-ANSI function declaration of function '%s'",
--
2.19.0
