What is the type of array pointer?

Question

When I declare an array in C:

char a[] = "something";

I understand that a is implicitly a const character pointer, i.e. 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?

a is not a pointer, it's an array. In some situations, it will "decay" to a pointer to a[0], or a char *... but &a does not give the address of a pointer, it gives the address of the array... which is the same address as &a[0], but a different type, as it points to an array and not to a char. char * const * const j is a pointer to a pointer, which &a is not. — Dmitri
– Dmitri, Commented Mar 22, 2017 at 0:05

flogram_dev · Accepted Answer · 2017-03-22 00:08:25Z

4

char a[] = "something";
I understand that a is implicitly a const character pointer, i.e. 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] (i.e. pointer to char[10]). char(*)[10] and char * const * are completely unrelated.

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

edited Mar 22, 2017 at 0:08

answered Mar 22, 2017 at 0:03

flogram_dev

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5 Comments

Akay Over a year ago

Then does char * const k = a; convert silently?

flogram_dev Over a year ago

Yes, that is known as array-to-pointer decay.

Akay Over a year ago

Oh..that explains. I thought array and pointer were internally represented the same way.

flogram_dev Over a year ago

Ah not at all. A pointer is represented as an integer, whereas an array is represented as a sequence of objects.

Akay Over a year ago

Hmmm...so after all the checks are done on compilation; at run time, are the array access operations just doing pointer arithmetic? Or is there something more to it? Or is that left up to the compiler?

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