c++ - 当 sizeof(long) == sizeof(int) 时 long* 和 int* 之间的转换

Question

我们通常知道 sizeof(long) != sizeof(int)。但是 (C++11) 标准的哪些部分不允许 long* 到 int* 别名？仅仅是 [conv.ptr] 中的遗漏，[basic.lval] 中的别名规则，还是其他原因？

void f()
{
    static_assert(sizeof(int) == sizeof(long), "");
    long x[] = {1, 2};
    int* y = x; // error: invalid conversion from ‘long int*’ to ‘int*’ [-fpermissive]
}

Answer 1

是的，在[conv.ptr]中被遗漏了适用段落在[expr.reinterpret.cast] ,

7 An object pointer can be explicitly converted to an object pointer of a different type. When a prvalue v of type "pointer to T1" is converted to the type "pointer to cv T2", the result is static_cast<cv T2*>(static_cast<cv void*>(v)) if both T1 and T2 are standard-layout types (3.9) and the alignment requirements of T2 are no stricter than those of T1, or if either type is void. Converting a prvalue of type "pointer to T1" to the type "pointer to T2" (where T1 and T2 are object types and where the alignment requirements of T2 are no stricter than those of T1) and back to its original type yields the original pointer value. The result of any other such pointer conversion is unspecified.

你必须使用 reinterpret_cast<int*>(...) .

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编辑。 我在这次编辑中使评论更加明显，这不是语言律师的问题，而是意图使用指针。我不确定这种担忧是如何产生的，因为很明显可以不问就简单地进行 C-cast，但如果有疑问 - 将指针转换为 int*违反strict aliasing规则。

也就是说，如果您打破了编译器关于不同类型的指针永远不会指向同一内存位置的假设，则可能会出现未定义的行为。