c: relax constexpr constraints, fix alignof

- Allow short-circuiting of constant logical expressions where the
  unevaluated operand is not a constant expression (`1 || 0/0`)

- Allow constant integer expressions that evaluate to zero to be used
  as null pointer constants (GNU extension).

    + According to the standard, `int *x = 5*0;` should be rejected.
      But compilers evaluate `5*0 -> 0` and allow it as if a null pointer
      literal.

1 files changed, 13 insertions, 0 deletions

diff --git a/test/17-misc.c b/test/17-misc.c
index ecff06c..521b948 100644
--- a/test/17-misc.c
+++ b/test/17-misc.c
@@ -13,6 +13,19 @@ int fn1(uvlong p_9) {
    return q;
 }
 
+/* Excerpt from linux/kernel.h */
+/*
+ * This returns a constant expression while determining if an argument is
+ * a constant expression, most importantly without evaluating the argument.
+ * Glory to Martin Uecker <Martin.Uecker@med.uni-goettingen.de>
+ */
+#define __is_constexpr(x) \
+	(sizeof(int) == sizeof(*(8 ? ((void *)((long)(x) * 0l)) : (int *)8)))
+
+_Static_assert(__is_constexpr(5 &&(1<<3) || 0/0), "");
+_Static_assert(!__is_constexpr(5/0), "");
+_Static_assert(!__is_constexpr(fn1(0)), "");
+
 extern int printf(const char *, ...);
 int main() {
    printf("%d\n", fn1(-77ull));