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[10/10] compiler.h: remove __is_constexpr()

Message ID 20241203-is_constexpr-refactor-v1-10-4e4cbaecc216@wanadoo.fr (mailing list archive)
State New
Headers show
Series compiler.h: refactor __is_constexpr() into is_const{,_true,_false}() | expand

Commit Message

Vincent Mailhol via B4 Relay Dec. 2, 2024, 5:33 p.m. UTC
From: Vincent Mailhol <mailhol.vincent@wanadoo.fr>

Now that all the users of __is_constexpr() have been migrated to
is_const() or one of its friends, remove it.

The homage to Martin Uecker's genius hack remains in the documentation
of __is_const_zero().

Signed-off-by: Vincent Mailhol <mailhol.vincent@wanadoo.fr>
---
 include/linux/compiler.h | 47 -----------------------------------------------
 1 file changed, 47 deletions(-)
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Patch

diff --git a/include/linux/compiler.h b/include/linux/compiler.h
index 165aa5b9bc484376087a130a1ac1f3edb50c983d..7ba75044828129cf1f81f458ade695786dbf132a 100644
--- a/include/linux/compiler.h
+++ b/include/linux/compiler.h
@@ -252,53 +252,6 @@  static inline void *offset_to_ptr(const int *off)
 #define __must_be_cstr(p) \
 	__BUILD_BUG_ON_ZERO_MSG(__annotated(p, nonstring), "must be cstr (NUL-terminated)")
 
-/*
- * 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>
- *
- * 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)))
-
 /*
  * Whether 'type' is a signed type or an unsigned type. Supports scalar types,
  * bool and also pointer types.