fix checking if type is void

Sparse warns if a void function returns a non-void expression.

But the check directly compares the returned type with void_ctype,
without taking in account the presence of a SYM_NODE. In
consequence, sparse issues a few false warnings.

Fix this by using is_void_type() to test the returned type.

Signed-off-by: Luc Van Oostenryck <luc.vanoostenryck@gmail.com>

1 files changed, 1 insertions, 1 deletions

diff --git a/evaluate.c b/evaluate.c
index dddea761..6d5bbca5 100644
--- a/evaluate.c
+++ b/evaluate.c
@@ -3544,7 +3544,7 @@ static struct symbol *evaluate_return_expression(struct statement *stmt)
 	fntype = current_fn->ctype.base_type;
 	rettype = fntype->ctype.base_type;
 	if (!rettype || rettype == &void_ctype) {
-		if (expr && expr->ctype != &void_ctype)
+		if (expr && !is_void_type(expr->ctype))
 			expression_error(expr, "return expression in %s function", rettype?"void":"typeless");
 		if (expr && Wreturn_void)
 			warning(stmt->pos, "returning void-valued expression");