In evaluate_assignment(), a local variable (named 'where') contains
the input expression (named 'expr', like in most other function).
This is doubly confusing because:
*) both variables hold the same pointer.
*) the name 'where' is normally used for a string with extra
information for error messages.
So, remove this intermediate and use the original 'expr' instead.
Signed-off-by: Luc Van Oostenryck <luc.vanoostenryck@xxxxxxxxx>
---
evaluate.c | 3 +--
1 file changed, 1 insertion(+), 2 deletions(-)
diff --git a/evaluate.c b/evaluate.c
index 947b121f4..53b66cc2d 100644
--- a/evaluate.c
+++ b/evaluate.c
@@ -1577,7 +1577,6 @@ static void evaluate_assign_to(struct expression *left, struct symbol *type)
static struct symbol *evaluate_assignment(struct expression *expr)
{
struct expression *left = expr->left;
- struct expression *where = expr;
struct symbol *ltype;
if (!lvalue_expression(left)) {
@@ -1591,7 +1590,7 @@ static struct symbol *evaluate_assignment(struct expression *expr)
if (!evaluate_assign_op(expr))
return NULL;
} else {
- if (!compatible_assignment_types(where, ltype, &expr->right, "assignment"))
+ if (!compatible_assignment_types(expr, ltype, &expr->right, "assignment"))
return NULL;
}
--
2.20.0
