constexpr: examine constness of conditionals at evaluation only

Move the whole calculation of conditional expressions' constness flags
to the evaluation phase such that expressions like

  0 ? __builtin_choose_expr(0, 0, 0) : 0
  0 ? 0 : __builtin_choose_expr(0, 0, 0)

can now be recognized as qualifying as integer constant expressions.

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

1 files changed, 2 insertions, 6 deletions

diff --git a/evaluate.c b/evaluate.c
index 07788eac..6d6e462c 100644
--- a/evaluate.c
+++ b/evaluate.c
@@ -1122,12 +1122,8 @@ static struct symbol *evaluate_conditional_expression(struct expression *expr)
 		true = &expr->cond_true;
 	}
 
-	if (expr->flags) {
-		int flags = expr->conditional->flags & CEF_ICE;
-		flags &= (*true)->flags & expr->cond_false->flags;
-		if (!flags)
-			expr->flags = CEF_NONE;
-	}
+	expr->flags = (expr->conditional->flags & (*true)->flags &
+			expr->cond_false->flags & ~CEF_CONST_MASK);
 
 	lclass = classify_type(ltype, &ltype);
 	rclass = classify_type(rtype, &rtype);