Question

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C: How come an array’s address is equal to its value?

SA
In C I tried to print the address of the pointer of an array.

int a[3] = {0,1,2};
printf("\n%p",a);
printf("\n%p",(&a));

the 2 statement prints the same value why?
thanks in advance

Answer 1

Any array argument to a function will decay to a pointer to the first element of the array.

The C Book has an excellent explanation of this:

Another case is when an array name is
the operand of the & address-of
operator. Here, it is converted into
the address of the whole array. What’s
the difference? Even if you think that
addresses would be in some way ‘the
same’, the critical difference is that
they have different types. For an
array of n elements of type T, then
the address of the first element has
type ‘pointer to T’; the address of
the whole array has type ‘pointer to
array of n elements of type T’;
clearly very different