indexing array in function from the pointer of some element

Question

Imagine some array

uint8_t var[5] = {1,2,3,4,5};

so var will be pointer to the first element of this array, and

uint 8_t* a=var;
b=a[3]

and

b=var[3]

will give the same result.

But will

a = &var[2];
b = a[1];

and

b=var[3];

be same?

Yes, pointer arithmetic is not limited to starting point at 0. — Jarod42
– Jarod42, Commented Jan 17, 2022 at 13:50
var is a uint8_t[5]. It can decay to a pointer to first element, but it doesnt always — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Jan 17, 2022 at 13:56
so var will be pointer No. var is an array. It takes 5*sizeof uint8_t bytes of memory. A pointer only takes sizeof uint8_t * bytes. You can use them in similar ways in most places but they are not the same. — Gerhardh
– Gerhardh, Commented Jan 17, 2022 at 14:04

Vlad from Moscow · Accepted Answer · 2022-01-17 14:02:47Z

3

After this assignment

a = &var[2];

that is the same as

a = var + 2;

due to the implicit conversion of the array designator to a pointer to its first element the pointer a points to the element var[2].

So a[0] yields var[2] and a[1] yields var[3].

Pay attention to that the subscript operator a[i] is evaluated like *( a + i ).

So you have a[1] is equivalent to *( a + 1 ) that is in turn equivalent to *( var + 2 + 1 ) that is to *( var + 3 ).

answered Jan 17, 2022 at 13:52

Vlad from Moscow

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1 Comment

And, even, a[-1] yields var[1].