数组指针的类型是什么？

Question

When I declare an array in C:

char a[] = "something";

I understand that a is implicitly a const character pointer, ie a is of the type: (char * const)

But then why does the following statement result in the compiler warning about incompatible pointer types?

char * const * j = &a;

The only way I've managed to get rid of it is by explicitly casting the right hand side to (char * const *) .

I hypothesized that & operator returns a constant pointer and tried:

char * const * const j = &a;

without success.

What is going on here?

Answer 1

 char a[] = "something"; 

I understand that a is implicitly a const character pointer, ie a is of the type: (char * const)

Wrong, a is of type (non-const) char[10] .

So now that we know a is char[10] , it's clear why the following doesn't compile:

char * const * j = &a;

&a is of type char(*)[10] (ie pointer to char[10] ). char(*)[10] and char * const * are completely unrelated.

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If you wrote char* a = "something"; , then a would be a char* .