14.11 Pointers and Arrays

The clean way to refer to an array element is array[index]. Another, complicated way to do the same job is to get the address of that element as a pointer, then dereference it: * (&array[0] + index) (or equivalently * (array + index)). This first gets a pointer to element zero, then increments it with + to point to the desired element, then gets the value from there.

That pointer-arithmetic construct is the definition of square brackets in C. a[b] means, by definition, *(a + b). This definition uses a and b symmetrically, so one must be a pointer and the other an integer; it does not matter which comes first.

Since indexing with square brackets is defined in terms of addition and dereferencing, that too is symmetrical. Thus, you can write 3[array] and it is equivalent to array[3]. However, it would be foolish to write 3[array], since it has no advantage and could confuse people who read the code.

It may seem like a discrepancy that the definition *(a + b) requires a pointer, while array[3] uses an array value instead. Why is this valid? The name of the array, when used by itself as an expression (other than in sizeof), stands for a pointer to the array’s zeroth element. Thus, array + 3 converts array implicitly to &array[0], and the result is a pointer to element 3, equivalent to &array[3].

Since square brackets are defined in terms of such an addition, array[3] first converts array to a pointer. That’s why it works to use an array directly in that construct.