排序时指针类型的身份转换是否视为资格转换
Is an identity conversion of pointer type be considered as qualification conversion when ranking
void fun(int*); // #1
void fun(int const*); //#2
int main(){
int* ptr = nullptr;
fun(ptr);
}
考虑以上示例，重载决议应 select #1 和 #2 中的最佳可行函数。这是this question的一个分支，即[over.ics.rank#3.2.1]和[over.ics.rank#3.2.5]之间的模糊部分。
相关规则为：
S1 is a proper subsequence of S2 (comparing the conversion sequences in the canonical form defined by [over.ics.scs], excluding any Lvalue Transformation; the identity conversion sequence is considered to be a subsequence of any non-identity conversion sequence) or, if not that
IMO，#1的S1有身份转换，而#2的S2有资格转换，上面的规则好像是说Identity conversion 优于任何具有完全匹配排名的非身份转换。
不过，还有一条相关的规则对应于这种情况，那就是：
over.ics.rank#3.2.5
S1 and S2 differ only in their qualification conversion ([conv.qual]) and yield similar types T1 and T2, respectively, where T1 can be converted to T2 by a qualification conversion.
IMO，这个规则好像说S1和S2都有一个资格转换但是这些转换是不同的T1可以通过限定转换转换为 T2。
在我的例子中，从int*到int const*绝对是一个资格转换。但是，一个edge-case是从int*到int*，这个转换算身份转换还是资格转换？根据conv.qual#3的定义，可以。不过严格来说应该叫身份转换.
问题 1：
我对规则[over.ics.rank#3.2.5]的理解对吗？或者，是否意味着，当T1可以通过限定转换转换为T2时，S1可能有一个空转换（例如身份转换）？ （包括指针的身份转换）
Q问题2：
指针类型的恒等转换是限定转换还是恒等转换？换句话说，[over.ics.rank#3.2.1] 或 [over.ics.rank#3.2.5] 是否适用于我的示例？
从 int* 到 int* 的标准转换序列不 包含限定转换。
[over.ics.best.general]/8 明确指出了这一点：
If no conversions are required to match an argument to a parameter type, the implicit conversion sequence is the standard conversion sequence consisting of the identity conversion ([over.ics.scs]).
从其他段落也可以推断。例如，参见 [over.best.ics.general]/1:
An implicit conversion sequence is a sequence of conversions used to convert an argument in a function call to the type of the corresponding parameter of the function being called.
The sequence of conversions is an implicit conversion as defined in [conv], which means it is governed by the rules for initialization of an object or reference by a single expression ([dcl.init], [dcl.init.ref]).
和[over.ics.best.general]/6:
When the parameter type is not a reference, the implicit conversion sequence models a copy-initialization of the parameter from the argument expression.
The implicit conversion sequence is the one required to convert the argument expression to a prvalue of the type of the parameter.
...
也就是说，int* -> int* 的正确隐式转换序列是根据语言规则，通常在复制 int* 时形成的序列- 从 int* 初始化。在这种情况下，[dcl.init.general]/16.9 适用：
Otherwise, the initial value of the object being initialized is the (possibly converted) value of the initializer expression.
A standard conversion sequence ([conv]) will be used, if necessary, to convert the initializer expression to the cv-unqualified version of the destination type; no user-defined conversions are considered.
If the conversion cannot be done, the initialization is ill-formed.
When initializing a bit-field with a value that it cannot represent, the resulting value of the bit-field is implementation-defined.
不需要转换，因为 int* 和 int* 已经是同一类型，所以不需要转换。它不是从 int* 到自身的限定转换。
这意味着[over.ics.rank]/3.2.1肯定是决定性的：S1是身份转换序列（不是由冗余限定转换组成的隐式转换序列）所以它是S2.
void fun(int*); // #1
void fun(int const*); //#2
int main(){
int* ptr = nullptr;
fun(ptr);
}
考虑以上示例，重载决议应 select #1 和 #2 中的最佳可行函数。这是this question的一个分支，即[over.ics.rank#3.2.1]和[over.ics.rank#3.2.5]之间的模糊部分。
相关规则为：
S1 is a proper subsequence of S2 (comparing the conversion sequences in the canonical form defined by [over.ics.scs], excluding any Lvalue Transformation; the identity conversion sequence is considered to be a subsequence of any non-identity conversion sequence) or, if not that
IMO，#1的S1有身份转换，而#2的S2有资格转换，上面的规则好像是说Identity conversion 优于任何具有完全匹配排名的非身份转换。
不过，还有一条相关的规则对应于这种情况，那就是：
over.ics.rank#3.2.5
S1 and S2 differ only in their qualification conversion ([conv.qual]) and yield similar types T1 and T2, respectively, where T1 can be converted to T2 by a qualification conversion.
IMO，这个规则好像说S1和S2都有一个资格转换但是这些转换是不同的T1可以通过限定转换转换为 T2。
在我的例子中，从int*到int const*绝对是一个资格转换。但是，一个edge-case是从int*到int*，这个转换算身份转换还是资格转换？根据conv.qual#3的定义，可以。不过严格来说应该叫身份转换.
问题 1：
我对规则[over.ics.rank#3.2.5]的理解对吗？或者，是否意味着，当T1可以通过限定转换转换为T2时，S1可能有一个空转换（例如身份转换）？ （包括指针的身份转换）
Q问题2：
指针类型的恒等转换是限定转换还是恒等转换？换句话说，[over.ics.rank#3.2.1] 或 [over.ics.rank#3.2.5] 是否适用于我的示例？
从 int* 到 int* 的标准转换序列不 包含限定转换。
[over.ics.best.general]/8 明确指出了这一点：
If no conversions are required to match an argument to a parameter type, the implicit conversion sequence is the standard conversion sequence consisting of the identity conversion ([over.ics.scs]).
从其他段落也可以推断。例如，参见 [over.best.ics.general]/1:
An implicit conversion sequence is a sequence of conversions used to convert an argument in a function call to the type of the corresponding parameter of the function being called. The sequence of conversions is an implicit conversion as defined in [conv], which means it is governed by the rules for initialization of an object or reference by a single expression ([dcl.init], [dcl.init.ref]).
和[over.ics.best.general]/6:
When the parameter type is not a reference, the implicit conversion sequence models a copy-initialization of the parameter from the argument expression. The implicit conversion sequence is the one required to convert the argument expression to a prvalue of the type of the parameter.
...
也就是说，int* -> int* 的正确隐式转换序列是根据语言规则，通常在复制 int* 时形成的序列- 从 int* 初始化。在这种情况下，[dcl.init.general]/16.9 适用：
Otherwise, the initial value of the object being initialized is the (possibly converted) value of the initializer expression. A standard conversion sequence ([conv]) will be used, if necessary, to convert the initializer expression to the cv-unqualified version of the destination type; no user-defined conversions are considered. If the conversion cannot be done, the initialization is ill-formed. When initializing a bit-field with a value that it cannot represent, the resulting value of the bit-field is implementation-defined.
不需要转换，因为 int* 和 int* 已经是同一类型，所以不需要转换。它不是从 int* 到自身的限定转换。
这意味着[over.ics.rank]/3.2.1肯定是决定性的：S1是身份转换序列（不是由冗余限定转换组成的隐式转换序列）所以它是S2.