"pointer to an object type"与"pointer to a void"比较相等时如何理解转换规则？
How to understand the conversion rule when a "pointer to an object type" compares for eqality with a "pointer to a void"?
n1570 6.5.9.5 (Equality operators) 说：
5 ......If one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void, the former is converted to the type of the latter.
如果"former"是"latter"是"pointer to object type"和"pointer to a void"，那么表示转换后比较发生在两个void*上，根据 6.5.9.2，未定义：
2.One of the following shall hold:
both operands have arithmetic type;
both operands are pointers to qualified or unqualified versions of compatible types;
one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void; or
one operand is a pointer and the other is a null pointer constant.
这是N1570的语言缺陷吗？
我没想过从这个角度，但我的理解是，两个void *是（转换后，根据相等运算符属性）兼容类型。
我参考的是
"- both operands are pointers to qualified or unqualified versions of compatible types;"
If "former" is "latter" are "pointer to object type" and "pointer to a void", then it means, after the conversion, the compare happens on two void*s,
是的，我确实是这样读的。
which according to 6.5.9.2, is undefined:
没有。您引用了该标准的第 6.5.9.2 节，显然是相信您所询问的案例未包含在该节的备选方案枚举列表中，但您错了。此备选方案涵盖案例：
one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void;
这里重要的是要理解指向void 的指针是指向对象类型的指针，对于
The void type comprises an empty set of values; it is an incomplete object type that cannot be completed.
(6.2.5/19；已强调)
6.5.9.2 不应被解释为暗示 void 不是对象类型；相反，它说 void * 与 all 对象指针类型相当，当然包括它本身。
正如@SouravGhosh 首先观察到的，此替代方案也涵盖了所讨论的情况：
both operands are pointers to qualified or unqualified versions of compatible types
因为每种类型都与自身兼容 (6.2.7/1)。
n1570 6.5.9.5 (Equality operators) 说：
5 ......If one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void, the former is converted to the type of the latter.
如果"former"是"latter"是"pointer to object type"和"pointer to a void"，那么表示转换后比较发生在两个void*上，根据 6.5.9.2，未定义：
2.One of the following shall hold:
both operands have arithmetic type;
both operands are pointers to qualified or unqualified versions of compatible types;
one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void; or
one operand is a pointer and the other is a null pointer constant.
这是N1570的语言缺陷吗？
我没想过从这个角度，但我的理解是，两个void *是（转换后，根据相等运算符属性）兼容类型。
我参考的是
"- both operands are pointers to qualified or unqualified versions of compatible types;"
If "former" is "latter" are "pointer to object type" and "pointer to a void", then it means, after the conversion, the compare happens on two
void*s,
是的，我确实是这样读的。
which according to 6.5.9.2, is undefined:
没有。您引用了该标准的第 6.5.9.2 节，显然是相信您所询问的案例未包含在该节的备选方案枚举列表中，但您错了。此备选方案涵盖案例：
one operand is a pointer to an object type and the other is a pointer to a qualified or unqualified version of void;
这里重要的是要理解指向void 的指针是指向对象类型的指针，对于
The
voidtype comprises an empty set of values; it is an incomplete object type that cannot be completed.
(6.2.5/19；已强调)
6.5.9.2 不应被解释为暗示 void 不是对象类型；相反，它说 void * 与 all 对象指针类型相当，当然包括它本身。
正如@SouravGhosh 首先观察到的，此替代方案也涵盖了所讨论的情况：
both operands are pointers to qualified or unqualified versions of compatible types
因为每种类型都与自身兼容 (6.2.7/1)。