在纯 C++ 中实现 Linux 内核的 __is_constexpr (ICE_P) 宏
Implementing the Linux Kernel's __is_constexpr (ICE_P) macro in pure C++
在 Martin Uecker 的 ICE_P 谓词的标准 C11 版本之后，我尝试用纯 C++ 实现它。 C11版本，使用_Generic选择如下：
#define ICE_P(x) _Generic((1? (void *) ((x)*0) : (int *) 0), int*: 1, void*: 0)
C++的明显做法是用模板替换_Generic和decltype，如：
template<typename T> struct is_ice_helper;
template<> struct is_ice_helper<void*> { enum { value = false }; };
template<> struct is_ice_helper<int*> { enum { value = true }; };
#define ICE_P(x) (is_ice_helper<decltype(1? (void *) ((x)*0) : (int *) 0)>::value)
然而，simplest test 失败了。为什么它不能检测整型常量表达式？
这个问题很微妙。确定条件表达式的指针操作数的复合类型的规范在 C++ 中与 C 中的类似，所以它开始看起来很有希望：
(N4659) [expr.cond]
7 Lvalue-to-rvalue, array-to-pointer, and function-to-pointer
standard conversions are performed on the second and third operands.
After those conversions, one of the following shall hold:
[...]
One or both of the second and third operands have pointer type; pointer conversions, function pointer conversions, and qualification
conversions are performed to bring them to their composite pointer
type (Clause [expr]). The result is of the composite pointer type.
[...]
还原为复合指针类型指定如下：
(N4659) [expr]
5 The composite pointer type of two operands p1 and p2 having
types T1 and T2, respectively, where at least one is a pointer or
pointer to member type or std::nullptr_t, is:
- if both p1 and p2 are null pointer constants,
std::nullptr_t;
- if either p1 or p2 is a null pointer constant, T2 or T1, respectively;
- if T1 or T2 is “pointer to cv1 void” and the other type is “pointer to cv2 T”, where T is an object type or void, “pointer to cv12 void”,
where cv12 is the union of cv1 and cv2;
- [...]
所以我们的 ICE_P 宏的结果取决于我们按顺序检查上面的哪一颗子弹。鉴于我们如何定义 is_ice_helper，我们知道复合类型不是 nullptr_t，否则我们会击中第一颗子弹，并且会由于缺少模板特化而出错。所以我们必须击中第 3 个项目符号，使谓词报告为假。这一切似乎都取决于空指针常量的定义。
(N4659) [conv.ptr] (emphasis mine)
1 A null pointer constant is an integer literal with value
zero or a prvalue of type std::nullptr_t. A null pointer
constant can be converted to a pointer type; the result is the null
pointer value of that type and is distinguishable from every other
value of object pointer or function pointer type. Such a conversion is
called a null pointer conversion. Two null pointer values of the same
type shall compare equal. The conversion of a null pointer constant to
a pointer to cv-qualified type is a single conversion, and not the
sequence of a pointer conversion followed by a qualification
conversion. A null pointer constant of integral type can be converted
to a prvalue of type std::nullptr_t.
由于根据上面的定义 (int*)0 不是空指针常量，我们不符合 [expr]/5 的第一个项目符号。复合类型是notstd::nullptr_t。 (void *) ((x)*0) 既不是空指针常量，也不能变成空指针常量。删除演员表（定义不允许的东西）给我们留下 (x)*0。这是一个值为零的整数常量表达式。但它不是零值的 整数文字 ！ C++中空指针常量的定义与C中的有偏差！
(N1570) 6.3.2.3 Pointers
3 An integer constant expression with the value 0, or such an
expression cast to type void *, is called a null pointer constant. If
a null pointer constant is converted to a pointer type, the resulting
pointer, called a null pointer, is guaranteed to compare unequal to a
pointer to any object or function.
C 允许任意 值为零的常量表达式构成空指针常量，而C++ 要求整型文字。鉴于 C++ 对计算各种文字类型的常量表达式的丰富支持，这似乎是一个不必要的限制。并且使上述 ICE_P 方法成为 C++ 中的非入门者。
在 ICE_P 谓词的标准 C11 版本之后，我尝试用纯 C++ 实现它。 C11版本，使用_Generic选择如下：
#define ICE_P(x) _Generic((1? (void *) ((x)*0) : (int *) 0), int*: 1, void*: 0)
C++的明显做法是用模板替换_Generic和decltype，如：
template<typename T> struct is_ice_helper;
template<> struct is_ice_helper<void*> { enum { value = false }; };
template<> struct is_ice_helper<int*> { enum { value = true }; };
#define ICE_P(x) (is_ice_helper<decltype(1? (void *) ((x)*0) : (int *) 0)>::value)
然而，simplest test 失败了。为什么它不能检测整型常量表达式？
这个问题很微妙。确定条件表达式的指针操作数的复合类型的规范在 C++ 中与 C 中的类似，所以它开始看起来很有希望：
(N4659) [expr.cond]
7 Lvalue-to-rvalue, array-to-pointer, and function-to-pointer standard conversions are performed on the second and third operands. After those conversions, one of the following shall hold:
[...]
One or both of the second and third operands have pointer type; pointer conversions, function pointer conversions, and qualification conversions are performed to bring them to their composite pointer type (Clause [expr]). The result is of the composite pointer type.
[...]
还原为复合指针类型指定如下：
(N4659) [expr]
5 The composite pointer type of two operands p1 and p2 having types T1 and T2, respectively, where at least one is a pointer or pointer to member type or
std::nullptr_t, is:
- if both p1 and p2 are null pointer constants,
std::nullptr_t;- if either p1 or p2 is a null pointer constant, T2 or T1, respectively;
- if T1 or T2 is “pointer to cv1 void” and the other type is “pointer to cv2 T”, where T is an object type or void, “pointer to cv12 void”, where cv12 is the union of cv1 and cv2;
- [...]
所以我们的 ICE_P 宏的结果取决于我们按顺序检查上面的哪一颗子弹。鉴于我们如何定义 is_ice_helper，我们知道复合类型不是 nullptr_t，否则我们会击中第一颗子弹，并且会由于缺少模板特化而出错。所以我们必须击中第 3 个项目符号，使谓词报告为假。这一切似乎都取决于空指针常量的定义。
(N4659) [conv.ptr] (emphasis mine)
1 A null pointer constant is an integer literal with value zero or a prvalue of type
std::nullptr_t. A null pointer constant can be converted to a pointer type; the result is the null pointer value of that type and is distinguishable from every other value of object pointer or function pointer type. Such a conversion is called a null pointer conversion. Two null pointer values of the same type shall compare equal. The conversion of a null pointer constant to a pointer to cv-qualified type is a single conversion, and not the sequence of a pointer conversion followed by a qualification conversion. A null pointer constant of integral type can be converted to a prvalue of typestd::nullptr_t.
由于根据上面的定义 (int*)0 不是空指针常量，我们不符合 [expr]/5 的第一个项目符号。复合类型是notstd::nullptr_t。 (void *) ((x)*0) 既不是空指针常量，也不能变成空指针常量。删除演员表（定义不允许的东西）给我们留下 (x)*0。这是一个值为零的整数常量表达式。但它不是零值的 整数文字 ！ C++中空指针常量的定义与C中的有偏差！
(N1570) 6.3.2.3 Pointers
3 An integer constant expression with the value 0, or such an expression cast to type
void *, is called a null pointer constant. If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.
C 允许任意 值为零的常量表达式构成空指针常量，而C++ 要求整型文字。鉴于 C++ 对计算各种文字类型的常量表达式的丰富支持，这似乎是一个不必要的限制。并且使上述 ICE_P 方法成为 C++ 中的非入门者。