Confusion between array and pointer variable

Question

we have

int a[2] = { 0, 1 } ; 
std::cout << &a ;
std::cout << a ;

Now when I run it, the output is the same for both!

Of what I have understood is that a decays to a pointer and gives me the address of the first element of the array i.e address of a[0].

Which is similar to printing std::cout << &a[0].

Now shouldn't std::cout << &a give me an error as a itself returns a pointer ?

No--the name of an array usually decays to a pointer--but not when used as the operand of the sizeof operator or the address-of operator (unary &). — Jerry Coffin
– Jerry Coffin, Commented Feb 16, 2015 at 5:47
It's worth pointing out that a and &a have different types, but both are convertible to void*. — Brian Bi
– Brian Bi, Commented Feb 16, 2015 at 5:48
There is overloaded operator<< that takes cout and a void *. All object pointers can be converted to void *, so there is no error. — M.M
– M.M, Commented Feb 16, 2015 at 5:52

Cheers and hth. - Alf · Accepted Answer · 2015-02-16 05:47:08Z

3

First, regarding

” a itself returns a pointer

No, the decay of array to pointer only happens in contexts where a pointer is expected. For example, it doesn't happen in a sizeof or decltype or typeid expression, or when the array is passed by reference to a function, or as here, when the address operator & is applied.

&a and &a[0] (or just a when decayed to pointer) refer to the same memory address, namely the first item of a, but have different types.

&a is a pointer to the whole array, and &a + 1 therefore points to a second such array following the first.

&a[0] (or just decayed a) is a pointer to the first item, and &a[0] + 1 therefore points to the second item.

answered Feb 16, 2015 at 5:47

Cheers and hth. - Alf

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