c++ - 如何理解 C++ 标准 N3337 中的 expr.const.cast 条款 8？

Question

在C++ draft standard N3337节[expr.const.cast]/8 :

The following rules define the process known as casting away constness. In these rules T_n and X_n represent types. For two pointer types:

X1 is T1 cv_1,1 * ⋯ cv_1,N * where T1 is not a pointer type

X2 is T2 cv_2,1 * ⋯ cv_2,M * where T2 is not a pointer type

K is min (N,M)

casting from X1 to X2 casts away constness if, for a non-pointer type T there does not exist an implicit conversion (Clause conv) from:

T cv_1,(N-K+1) * cv_1,(N-K+2) *⋯ cv_1,N *

to

T cv_2,(M-K+1) * cv_2,(M-K+2) *⋯ cv_2,M *

从句看不懂，于是在C++ draft standard N4659中查找对应的从句, [expr.const.cast]/7 :

A conversion from a type T1 to a type T2 casts away constness if T1 and T2 are different, there is a cv-decomposition of T1 yielding n such that T2 has a cv-decomposition of the form

cv₀²P₀² cv₁²P₁² ⋯ cv_n-1²P_n-1² cv_n² U₂

and there is no qualification conversion that converts T1 to

cv₀²P₀¹ cv₁²P₁¹ ⋯ cv_n-1²P_n-1¹ cv_n² U₁.

但是我还是搞不懂T1,T2和T之间的关系。帮帮我，谁能解释一下这个子句？

Answer 1

在旧版本中，引入了一个(大部分)任意的 T 只是为了免除

const int *p=0;
auto q=reinterpret_cast<char*>(p);  // error: casts away constness

正在从 int 更改为 char。它被有效地重写为

auto __q=reinterpret_cast<const char*>(p);  // ok
char *q=__q;  // error

较新的版本只是明确地重用了 U1 而不是发明了 T。