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The clean way to refer to an array element is

. Another, complicated way to do the
same job is to get the address of that element as a pointer, then
dereference it: `array`[`index`]`* (&`

(or
equivalently `array`[0] + `index`)`* (`

). This first gets a
pointer to element zero, then increments it with `array` + `index`)`+`

to point to
the desired element, then gets the value from there.

That pointer-arithmetic construct is the *definition* of square
brackets in C.

means, by definition,
`a`[`b`]`*(`

. This definition uses `a` + `b`)`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 dereference, 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 `*(`

requires a pointer, but `a` +
`b`)`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 arrays’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 addition,
`array[3]`

first converts `array`

to a pointer. That’s why
it works to use an array directly in that construct.

Next: Pointer Arithmetic Low Level, Previous: Pointer Arithmetic, Up: Pointers [Contents][Index]