Why does x[y] == y[x] in c++?

Question

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

Someone told me this... I didn't believe them at first but it does work. If x and y do not change throughout the code, why does this work:

int x [5] = { 0,1,2,3,4};
int y = 3;

if(x[y] == y[x]){
    cout << "Why..." << endl;
}

How does x array's value in index y is = the x index's value's in array y? But there was no y array.

This has been asked and answered gazillion times in various FAQs as well as here, on SO. C FAQ link: http://c-faq.com/aryptr/joke.html — AnT stands with Russia, Nov 18 '11 at 16:27

score 6 · Answer 1 · answered Nov 18 '11 at 16:26

6

It is always true (for normal operator==)

a[i]  --> *(a+i) --> *(i+a) --> i[a]

since int is intrinsic and has commutative operator==, this will always be true

answered Nov 18 '11 at 16:26

sehe

@delnan: How was that not the question? Both expressions are pointer dereference of the same pointer addition. – Ben Voigt Nov 18 '11 at 16:29
2

More importantly, because `int*` doesn't have an overloaded `operator[]`! – Kerrek SB Nov 18 '11 at 16:29

score 3 · Answer 2 · answered Nov 18 '11 at 16:27

3

Because all of the following are same:

x[y] == y[x] == *(x+y) == *(y+x)

answered Nov 18 '11 at 16:27

Alok Save

score 2 · Answer 3 · answered Nov 18 '11 at 16:27

2

Because x[y] is just another way to say *(x + y), and that is the same as *(y + x).

answered Nov 18 '11 at 16:27

Paul Manta

You mean "...another way to say `*(x + y)`, and that is the same as `*(y + x)`. – Martin B Nov 18 '11 at 16:29
No, it's saying `*(x + y)`. `&x[y]` would be saying `x + y`. – Chris Lutz Nov 18 '11 at 16:30

3 Answers3