Const 转换为非指针非引用类型

Question

来自[expr.const.cast]/3：

For two similar types T1 and T2, a prvalue of type T1 may be explicitly converted to the type T2 using a const_cast if, considering the cv-decompositions of both types, each P_i¹ is the same as P_i² for all i. The result of a const_cast refers to the original entity.

似乎允许将 const 转换为非指针非引用类型。比如下面的函数

void f(int a)
{
    const_cast<int>(a);
}

应该是合式的，因为 int 和 int 肯定是相似的类型并且没有 P_i 在他们的 cv 分解中（因此命题 "each P_i¹ is the same as P_i² for all i" 应该是真的）。

但是，GCC 和 Clang 都拒绝上面的代码（参见 Compiler Explorer）。错误信息是

叮当声：

<source>: In function 'void f(int)':
<source>:3:22: error: invalid use of const_cast with type 'int', which is not a pointer, reference, nor a pointer-to-data-member type
    3 |     const_cast<int>(a);
      |                      ^

海湾合作委员会：

<source>: In function 'void f(int)':
<source>:3:5: error: invalid use of 'const_cast' with type 'int', which is not a pointer, reference, nor a pointer-to-data-member type
    3 |     const_cast<int>(a);
      |     ^~~~~~~~~~~~~~~~~~

我是不是遗漏了什么或者是编译器错误？

更新：这也行不通：

void f()
{
    const_cast<int>(int{});
}

Answer 1

在 C++17 中，类型不相似，因此引用的文本不适用。所以这个 const_cast 是不允许的，因为除非明确允许，否则不允许 const_cast。

C++17 [conv.qual]/1:

A cv-decomposition of a type T is a sequence of cv_i and P_i such that T is

“cv₀ P₀ cv₁ P₁ ··· cv_n-1 P_n-1 cv_n U” for n > 0,

where each cv_i is a set of cv-qualifiers (6.9.3), and each P_i is “pointer to” (11.3.1), “pointer to member of class C_i of type” (11.3.3), “array of N_i ”, or “array of unknown bound of” (11.3.4). [...]

然后

Two types T₁ and T₂ are similar if they have cv-decompositions with the same n such that corresponding P_i components are the same and the types denoted by U are the same.

要求n > 0表示必须有cv₀ P₀,即类型中至少有一个指针。

由于 Issue 2051，最新的 C++20 草案将 n > 0 更改为 n ≥ 0 .但不改变 const_cast 的规范。我不能说这是有意还是疏忽。

因此 C++20 可能会使您的 const_cast 表达式定义良好，编译器将不得不跟上。

Const 转换为非指针非引用类型

Const cast to non-pointer non-reference type

c++

language-lawyer

const-cast