C array arithmetic and pointers

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In C arrays why is this true? a[5] == 5[a]

I am reading through a tutorial on C and I came across this syntax:

int doses[] = {1, 3, 2, 1000};
doses[3] == *(doses + 3) == *(3 + doses) == 3[doses]

Now the point is to get the int 1000, but the last one doesn't make any sense. Either its late and my brain is not functioning, its something specific to C, or its a typo. I want to cover all my basics when it comes to pointers so reading through it carefully. That means understanding it all. Any answers would be much appreciated!

score 7 · Accepted Answer · answered Dec 02 '12 at 09:37

7

From Wikipedia

Since the expression a[i] is semantically equivalent to *(a+i), which in turn is equivalent to *(i+a), the expression can also be written as i[a], although this form is rarely used.

answered Dec 02 '12 at 09:37

Rotem

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"*... as i[a], although this form is rarely used.*": Yes, very rarely. – alk Dec 02 '12 at 09:37
Wow, I really appreciate the timely response. That just made no sense to me coming from Java and PHP, but semantically I guess I can live with that. – Andy Dec 02 '12 at 09:39
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@alk it's still good enough for obfuscation, though. – John Dvorak Dec 02 '12 at 09:39
@Andy I'll be honest with you, I had no idea either. I googled "c array syntax" and this was the first result. – Rotem Dec 02 '12 at 09:39
@alk also note the syntax `&(a[b])` which is equivalent to `a+b` and also works for non-pointers. – John Dvorak Dec 02 '12 at 09:40
Really? Haha. I didn't think I'd get anything. Thats embarrassing. But hey, now its an actual question on here :D – Andy Dec 02 '12 at 09:41
@Andy I agree, it's an interesting question! – Rotem Dec 02 '12 at 09:44

score 2 · Answer 2 · answered Dec 02 '12 at 09:39

2

Yes, array subscripting is commutative in C. e1[e2] is indeed the same as *((e1)+(e2)). But it is useless in production code, and the only purpose of this notation is to make obfuscated source code.

answered Dec 02 '12 at 09:39

md5

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I'm curious if you know the reason as to why this decision was made in the design of array/pointer arithmetic? I agree it seems useless and not to mention confusing. But I guess now that I know it wouldn't be so bad. – Andy Dec 02 '12 at 09:47
@Andy: When C programming langage was designed, compilers didn't have enough memory to perform a lot of syntax check. So translating `E1[E2]` to `*((E1)+(E2))` reduced the cost of the compilation. – md5 Dec 02 '12 at 09:58

Omkant · Answer 3 · 2012-12-02T10:02:04.203

ISO c99 : 6.5.2.1 Array subscripting

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One of the expressions shall have type ‘‘pointer to object type’’, the other expression shall have integer type, and the result has type ‘‘type’’.

E1[E2] either E1 will be pointer to object type and E2 will be integer type.

Or,

E1 is integer type and E2 is of pointer to that type

as + is commutative so E1[E2] == E2[E1] , because it's actually evaluated as (*(E1+E2))

C array arithmetic and pointers

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