array[2] == 2[array], why is that?

[mattmao]

Hello everyone:

Recently we were given this example:

int array[10]={0,1,2,3,4,5,6,7,8,9};

Of course if you printf("%d", array[2]); the output would be 2.

But our tutor raised such a confusing debate:

in math x + 2 = 2 + x.
and we knew array is but a pointer to int.
therefore array[2] means the location of the address of array plus two memory unit for int.

Thus

array[2] = array +2 = 2 + array = 2[array]

Given this formula, we were astonished to see the output of
printf("%d", 2[array]) is identical to the previous one.


We accepted this but at least for me, I really don't know why...

[Ganon11]

Umm...wow. That's surprising, but it makes sense. So array is really just an address in memory, for sake of argument let's call this address 282. Saying array[2] is like saying "Go to the address pointed to by array, go 2 slots further, and use the information there." Substituting in the value of array, you get the memory location 284.

Reversing that, you take the memory location 2 and add the address pointed to by array, and you get 284.

I guess.

[mattmao]

This makes fully sense.

However, wouldn't the OS block you from getting the address of "2"?

Or is it true that the Os automatically reverse the process and locate the address of array in memory prior to shifting two units?

If that is the case, then this rare expression would be a trick for us.

[shabinesh]

hey,
array+2 and 2+aray is not same as array[2] its a mistake.It should be like this
*(array+2) and *(2+array) respectively.the internal representation of array is converted as *(array+2) so array[2] and 2[array] are same.

[Banfa]

However, wouldn't the OS block you from getting the address of "2"?

It doen't try to get the address of 2, it uses 2 as a number in calculating the required address.