When an expression that needs to be constant but is, in fact,
not constant, sparse throws an error and leaves it as-is.
But some code makes the assumption that the expression is
constant and uses its value, with some random result.
One situation where this happens is in code like:
switch (x) {
case <some non-const expression>: ...
In this case, the linearization of the switch/case statement
will unconditionally use the value of the case expression
but the expression has no value.
One way to avoid this would be to add defensive checks each
time a value is retrieved but this is a lot of work and time
for no benefits.
So, change this by forcing the expression to be a constant
value of 0 just after the error message has been issued.
Signed-off-by: Luc Van Oostenryck <luc.vanoostenryck@xxxxxxxxx>
---
expand.c | 6 +++++-
1 file changed, 5 insertions(+), 1 deletion(-)
diff --git a/expand.c b/expand.c
index 623b180025ad..c4f806dee1ba 100644
--- a/expand.c
+++ b/expand.c
@@ -1177,8 +1177,12 @@ static void expand_const_expression(struct expression *expr, const char *where)
{
if (expr) {
expand_expression(expr);
- if (expr->type != EXPR_VALUE)
+ if (expr->type != EXPR_VALUE) {
expression_error(expr, "Expected constant expression in %s", where);
+ expr->ctype = &int_ctype;
+ expr->type = EXPR_VALUE;
+ expr->value = 0;
+ }
}
}
--
2.28.0
