RE: [PATCH v4 1/2] compiler.h: add const_true()

[David Laight]

From: Vincent Mailhol
> Sent: 13 November 2024 17:19
> 
> __builtin_constant_p() is known for not always being able to produce
> constant expression [1] which led to the introduction of
> __is_constexpr() [2]. Because of its dependency on
> __builtin_constant_p(), statically_true() suffers from the same
> issues.

Chalk and cheese.
Personally I don't think any of the text below is really needed.

You might want the final short form - but it is a short form.

OTOH the implementation is horrid.

You probably want to start with this (posted a while back in a minmax patch set:

diff --git a/include/linux/compiler.h b/include/linux/compiler.h
index 2594553bb30b..35d5b2fa4786 100644
--- a/include/linux/compiler.h
+++ b/include/linux/compiler.h
@@ -242,6 +242,23 @@ static inline void *offset_to_ptr(const int *off)
 /* &a[0] degrades to a pointer: a different type from an array */
 #define __must_be_array(a)     BUILD_BUG_ON_ZERO(__same_type((a), &(a)[0]))

+/**
+ * __if_constexpr - Check whether an expression is an 'integer
+ *             constant expression'
+ * @expr: Expression to test, not evaluated, can be a pointer
+ * @if_const: return value if constant
+ * @if_not_const: return value if not constant
+ *
+ * The return values @if_const and @if_not_const can have different types.
+ *
+ * Relies on typeof(x ? NULL : ptr_type) being ptr_type and
+ * typeof(x ? (void *)y : ptr_type) being 'void *'.
+ */
+#define __if_constexpr(expr, if_const, if_not_const)           \
+       _Generic(0 ? ((void *)((long)(expr) * 0l)) : (char *)0, \
+               char *: (if_const),                             \
+               void *: (if_not_const))
+
 /*
  * This returns a constant expression while determining if an argument is
  * a constant expression, most importantly without evaluating the argument.
--

Then have:

#define __is_constexpr(x) __if_constexpr(x, 1, 0)

#define const_true(x) __if_constexpr(x, x, 0)

	David

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