问题描述
据我了解，这样的事情还可以:
It is my understanding that something like this is okay:
const int ci = 42;
const int *cip = &ci;
int *ip = (int *)cip;
int j = *ip;
那呢?
const int ci = 42;
const int *cip = &ci;
const int **cipp = &cip;
int **ipp = (int **)cipp;
int j = **ipp;
推荐答案
表达式*ipp是类型为int *的左值，但是它被用于访问有效类型为const int *的对象. (即cip).
The expression *ipp is an lvalue of type int *, however it is being used to access an object of effective type const int *. (Namely, cip).
根据标准的字母，这是对别名的严格违反:允许的别名类型列表不包含别名T *的别名为const T *，反之亦然.
According to the letter of the standard, it is a strict aliasing violation: the list of allowed types to alias does not include aliasing T * as const T * or vice versa.
最接近的例外情况是:(C11 6.5/6摘录)
The closest exception is this one: (C11 6.5/6 excerpt)
合格版本"由C11 6.2.5/26明确定义:
"qualified version" is clearly defined by C11 6.2.5/26:
因此，例外情况是T可以别名为const T，反之亦然，但是指向可别名类型的指针没有类似的例外. const T *不是T *的限定版本.
So the exception is that T may be aliased as const T and vice versa, but there is no similar exception for pointers to aliasable types. const T * is not a qualified version of T *.
但是当然有脚注:
我不能说规则的意图是否使const T *和T *具有别名.对我来说，不清楚T *和const T *具有相同的表示和对齐要求"(6.2.5/28)的目的是什么，如果它不是别名的话.
I couldn't say whether the intent of the rule is for const T * and T * to be aliasable or not. It seems unclear to me what the purpose of specifying that T * and const T * have "the same representation and alignment requirements" (6.2.5/28) would be if it is not aliasable.
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