问题描述
一个指针非const数据可以隐式转换的指针相同类型的常量数据
A pointer to non-const data can be implicitly converted to a pointer to const data of the same type:
int *x = NULL;
int const *y = x;
添加额外的常量预选赛相匹配的额外间接理应以同样的方式：
Adding additional const qualifiers to match the additional indirection should logically work the same way:
int * *x = NULL;
int *const *y = x; /* okay */
int const *const *z = y; /* warning */
与海湾合作委员会或锵与 -Wall 标志进行编译，但是，这会导致以下警告：
Compiling this with GCC or Clang with the -Wall flag, however, results in the following warning:
test.c:4:23: warning: initializing 'int const *const *' with an expression of type
'int *const *' discards qualifiers in nested pointer types
int const *const *z = y; /* warning */
^ ~
为什么增加一个额外的常量预选赛放弃嵌套指针类型的限定？
Why does adding an additional const qualifier "discard qualifiers in nested pointer types"?
推荐答案
之所以常量只能添加一个层次深的是潜移默化的，是由问题11.10在comp.lang.c常见问题解答。
The reason why const can only be added one level deep is subtle, and is explained by Question 11.10 in the comp.lang.c FAQ.
简单地说，考虑这个例子密切相关，你的：
Briefly, consider this example closely related to yours:
const int i;
int *p;
int const **z = &p;
*z = &i;
/* Now p points to i */
下用只允许分配到丢弃预选赛第一指向的级别（所以分配到以Z 在这里是不允许的）。
您确切的例子不存在这个问题，因为常量第二个层次是指分配给 * Z 将不会被允许反正。 C ++的会的允许它在这种情况下，准确，但C'S简单的规则不你的情况和上面的例子加以区分。
Your exact example does not suffer from this problem, because the const the second level means that the assignment to *z would not be allowed anyway. C++ would allow it in this exact case, but C's simpler rules do not distinguish between your case and the example above.
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