c++ - const强制转换为非指针非引用类型

Question

从[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const强制转换为非指针非引用类型是允许的。例如，以下功能

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

应该格式正确，因为 int和 int肯定是相似的类型，并且在cv分解中没有Pi(因此“每个Pi1与所有i的Pi2相同”的主张应该成立)。

但是，GCC和Clang都拒绝上面的代码(请参阅 Compiler Explorer)。错误消息是

铛:

<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);
      |                      ^

GCC:

<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);
      |     ^~~~~~~~~~~~~~~~~~

我是否缺少某些内容？或者它是编译器错误？

UPDATE :这不起作用，或者:

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

Answer 1

在C++ 17中，类型不同，因此引用的文本不适用。因此，不允许此const_cast，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 意味着必须存在cv0 P0，即该类型中至少有一个指针。

最新的C++ 20草案由于 Issue 2051将 n> 0 更改为 n≥0 。但是不更改 const_cast的规范。我不能说这是故意还是疏忽。

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