c++ - 如何使用包含的转换的等级比较两个标准转换序列

Question

#include <iostream>
void g(int*);  //#1
void g(int (&arr)[2]);  //#2

void f(int*);  //#3
void f(int const*);  //#4
int main(){
  int arr[2] ={0};
  f(arr);    // choose #3
  g(arr);  //ambiguous
}

考虑上面的代码， #3 被选为 f(ptr)然而， g(arr)给出 ambiguous诊断。
选择最佳函数的规则定义为:

Standard conversion sequence S1 is a better conversion sequence than standard conversion sequence S2 if

• 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

所以看看 over.ics.scs#3

These are used to rank standard conversion sequences. The rank of a conversion sequence is determined by considering the rank of each conversion in the sequence and the rank of any reference binding.

根据我对上述规则的理解，可以理解为什么 #3是 f(ptr) 的最佳过载， 那是:
将 S1 设为 (arr => int*):

Array-to-pointer conversion -> (identity conversion)  
^^^^^^^^^^^^^^^^^^^^^^^^^^^    ^^^^^^^^^^^^^^^^^^^^^                   
     int[2] => int*             int* => int* 

同时将 S2 设为 (ptr => int const*)

Array-to-pointer conversion -> Qualification conversions ->  identity conversion   
^^^^^^^^^^^^^^^^^^^^^^^^^^^    ^^^^^^^^^^^^^^^^^^^^^^^^^     ^^^^^^^^^^^^^^^^^^^ 
     int[2] => int*               int* => int const*           int const* => int const* 

自 identity conversion是 Qualification conversions 的真子序列，因此 S1 优于 S2。所以， #3 由 f(ptr) 的重载决议选择.
当我使用类似的过程来确定哪个最适合 g(arr) ，我遇到了一个问题。
同样，给定 S1 为 (arr => int*)

Array-to-pointer conversion -> identity conversion  
^^^^^^^^^^^^^^^^^^^^^^^^^^^    ^^^^^^^^^^^^^^^^^^^ 
      int[2] => int*              int* => int*

而给定 S2 as(arr => int (&arr)[2])

When a parameter of reference type binds directly to an argument expression, the implicit conversion sequence is the identity conversion, unless the argument expression has a type that is a derived class of the parameter type, in which case the implicit conversion sequence is a derived-to-base Conversion

identity conversion
^^^^^^^^^^^^^^^^^^^
  bind to reference   

在这里， identity conversion的 S2是 Array-to-pointer conversion 的真子序列的 S1 ，因此它应该比 S1 更好，为什么编译器会提示 g(arr)是模棱两可的调用吗？
我对如何对标准转换序列进行排名有任何误读吗？如何比较两个标准 ICS(包含转换的等级)？

Answer 1

关键点在这里:

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

这意味着，对于函数调用 g(arr) , 全部 数组到指针的转换 不用于确定排名。换句话说，来自类型 int[2]输入 int* ，只有一个身份转换用于确定排名。因此， void g(int*); 的 S1和 void g(int (&arr)[2]); 的 S2是无法区分的 ICS，因此编译器给出了一个不明确的错误。
相比之下， void f(int*); 的转换和 void f(int const*);用于比较排名的是 identity conversion和 qualification conversion ， 分别。
根据规则:

the identity conversion sequence is considered to be a subsequence of any non-identity conversion sequence

因此， Qualification conversion被认为比 identity conversion 的排名更差.所以， void f(int*)赢得了比赛。