C weird array syntax in multi-dimensional arrays

Question

I've known that this is true:

x[4] == 4[x]

What is the equivalent for multi-dimensional arrays? Is the following true?

x[4][3] == 3[x[4]] == 3[4[x]]

@JimClay Because `x[4] == *(x + 4)` so `4[x] == *(4 + x)`. The `[]` operator is just syntactic sugar. — Andrew Marshall, Nov 13 '11 at 02:55
Thanks for the explanations. I'm both fascinated and repelled at once. — Jim Clay, Nov 13 '11 at 02:57
@JimClay: ..and that's what C is really for. repelling and fascinating. — kbyrd, Nov 13 '11 at 02:59
@Carl: Yes, sorry for offering a question as silly as that. I just had no compiler at this pc. Just tested online and it works. Pubby's explanation is more than enough, too. — ypercubeᵀᴹ, Nov 13 '11 at 03:07

Pubby · Accepted Answer · 2011-11-13T02:59:28.883

15

x[y] is defined as *(x + (y))

x[y][z] would become *(*(x + (y)) + z)

x[y[z]] would become *(x + (*(y + (z))))

x[4][3] would become *(*(x + (4)) + 3) would become *(*(x + 4) + 3)

3[x[4]] would become *(3 + (*(x + (4)))) would become *(*(x + 4) + 3)

3[4[x]] would become *(3 + (*(4 + (x)))) would become *(*(x + 4) + 3)

Which means they are all equivalent.

edited Nov 13 '11 at 02:59

answered Nov 13 '11 at 02:54

Pubby

In the case of a multidimensional array `int x[5][7]` you'd have `x[y][z]` defined as `*(x + 7*y + z)`. And `x[y]` would become `x + 7*y` i.e. the pointer to the indicated slice. But I believe the equivalences still hold, even if the expanded expressions are somewhat longer. If the compiler does accept the code at all, that is. – MvG Jul 31 '12 at 23:16

score 1 · Answer 2 · answered Nov 13 '11 at 02:55

1

Yes. In each case x is an array which decays to a pointer and then has pointer arithmetic performed on it.

answered Nov 13 '11 at 02:55

Mankarse

2 Answers2