[PATCH] compiler.h: Explain how __is_constexpr() works

From: Kees Cook
Date: Thu Feb 29 2024 - 23:44:55 EST


The __is_constexpr() macro is dark magic. Shed some light on it with
a comment to explain how and why it works.

Acked-by: Gustavo A. R. Silva <gustavoars@xxxxxxxxxx>
Signed-off-by: Kees Cook <keescook@xxxxxxxxxxxx>
---
Cc: Rasmus Villemoes <linux@xxxxxxxxxxxxxxxxxx>
Cc: Miguel Ojeda <miguel.ojeda.sandonis@xxxxxxxxx>
Cc: Jani Nikula <jani.nikula@xxxxxxxxxxxxxxx>
Cc: David Laight <David.Laight@xxxxxxxxxx>
Cc: Nick Desaulniers <ndesaulniers@xxxxxxxxxx>
Cc: Martin Uecker <Martin.Uecker@xxxxxxxxxxxxxxxxxxxxx>
Cc: Jonathan Corbet <corbet@xxxxxxx>
Cc: linux-doc@xxxxxxxxxxxxxxx
v2: *thread necromancy* rewrite based on feedback to v1
v1: https://lore.kernel.org/all/20220131204357.1133674-1-keescook@xxxxxxxxxxxx/
---
include/linux/compiler.h | 39 +++++++++++++++++++++++++++++++++++++++
1 file changed, 39 insertions(+)

diff --git a/include/linux/compiler.h b/include/linux/compiler.h
index bb1339c7057b..38cd9f3c8f6a 100644
--- a/include/linux/compiler.h
+++ b/include/linux/compiler.h
@@ -231,6 +231,45 @@ static inline void *offset_to_ptr(const int *off)
* 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@xxxxxxxxxxxxxxxxxxxxx>
+ *
+ * Details:
+ * - sizeof() return an integer constant expression, and does not evaluate
+ * the value of its operand; it only examines the type of its operand.
+ * - The results of comparing two integer constant expressions is also
+ * an integer constant expression.
+ * - The first literal "8" isn't important. It could be any literal value.
+ * - The second literal "8" is to avoid warnings about unaligned pointers;
+ * this could otherwise just be "1".
+ * - (long)(x) is used to avoid warnings about 64-bit types on 32-bit
+ * architectures.
+ * - The C Standard defines "null pointer constant", "(void *)0", as
+ * distinct from other void pointers.
+ * - If (x) is an integer constant expression, then the "* 0l" resolves
+ * it into an integer constant expression of value 0. Since it is cast to
+ * "void *", this makes the second operand a null pointer constant.
+ * - If (x) is not an integer constant expression, then the second operand
+ * resolves to a void pointer (but not a null pointer constant: the value
+ * is not an integer constant 0).
+ * - The conditional operator's third operand, "(int *)8", is an object
+ * pointer (to type "int").
+ * - The behavior (including the return type) of the conditional operator
+ * ("operand1 ? operand2 : operand3") depends on the kind of expressions
+ * given for the second and third operands. This is the central mechanism
+ * of the macro:
+ * - When one operand is a null pointer constant (i.e. when x is an integer
+ * constant expression) and the other is an object pointer (i.e. our
+ * third operand), the conditional operator returns the type of the
+ * object pointer operand (i.e. "int *). Here, within the sizeof(), we
+ * would then get:
+ * sizeof(*((int *)(...)) == sizeof(int) == 4
+ * - When one operand is a void pointer (i.e. when x is not an integer
+ * constant expression) and the other is an object pointer (i.e. our
+ * third operand), the conditional operator returns a "void *" type.
+ * Here, within the sizeof(), we would then get:
+ * sizeof(*((void *)(...)) == sizeof(void) == 1
+ * - The equality comparison to "sizeof(int)" therefore depends on (x):
+ * sizeof(int) == sizeof(int) (x) was a constant expression
+ * sizeof(int) != sizeof(void) (x) was not a constant expression
*/
#define __is_constexpr(x) \
(sizeof(int) == sizeof(*(8 ? ((void *)((long)(x) * 0l)) : (int *)8)))
--
2.34.1