c - "signed or unsigned type"在此 C90 未定义行为定义中的含义是什么？

Question

在 ANSI C90 标准中，第 6.3 节对表达式有如下说明:

An object shall have its stored value accessed only by an lvalue that has one of the following types: [...] a type that is the signed or unsigned type corresponding to a qualified version of the declared type of the object

附件 G.2 中有这个未定义行为的实例:

The behavior in the following circumstances is undefined: [...] An object has its stored value accessed by an lvalue that does not have one of the following types: the declared type of the object, a qualified version of the declared type of the object, the signed or unsigned type corresponding to the declared type of the object, the signed or unsigned type corresponding to a qualified version of the declared type of the object, an aggregate or union type that (recursively) includes one of the aforementioned types among its members, or a character type (6.3).

我发现强调部分的措辞含糊不清，很难解释。

1. 是否表示“有符号则为原类型对应的有符号类型，无符号则为原类型对应的无符号类型”；或“原始类型对应的类型(有符号或无符号无关紧要)”？也就是说，是:

   signed int a = -10;
   unsigned int b = *((unsigned int *) a);
   

   ...未定义？

2. 如果有符号/无符号无关紧要，因为标准区分了三种类型 char、signed char 和 unsigned char，是否会定义通过 signed char * 或 unsigned char * 访问 char？

Answer 1

它是说将值转换为不同的符号不是不是未定义的行为。如果对象声明为 signed int，您可以使用 unsigned int 左值访问它，反之亦然。

当它说“对象的声明类型”时，已经涵盖了签名相同的情况，尽管这种情况也可以被认为是这样说。

在 char 的情况下，signed char 和 unsigned char 都是“对应于”该类型的“有符号或无符号类型”。

总而言之，这只是说左值的符号性不会影响访问是否明确定义。