A 'Strange Phenomenon' in C/C++ Arrays

Everyone is familiar with using arrays, right?
Take a look at this program, it’s quite simple.

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#include<iostream>
int main()
{
    int a[] = {1,2,3,4,5};
    for(int i = 0 ; i < 5; i++)
        std::cout << i[a] << " ";
    return 0;
}

Now look carefully at line 6.
What did you notice?
Try compiling it to see if it passes?

Let’s simplify this program even more

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int a[5] = {1,2,3,4,5};
int b = 1[a];

Now let’s look at the generated assembly code

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4:       int a[5] = {1,2,3,4,5};
00401568   mov         dword ptr [ebp-14h],1
0040156F   mov         dword ptr [ebp-10h],2
00401576   mov         dword ptr [ebp-0Ch],3
0040157D   mov         dword ptr [ebp-8],4
00401584   mov         dword ptr [ebp-4],5
5:        int b = 1[a];
0040158B   mov         eax,dword ptr [ebp-10h]
0040158E   mov         dword ptr [ebp-18h],eax

You’re not seeing things wrong. At this moment, this array is possessed by Chuck Norris - the addresses pointed to by a[1] and 1[a] are the same, both are [ebp-10h].

Why?

Let’s recall the relationship between arrays and pointers. How do we represent arrays with pointers?
*a is a reference to the value at index 0 in array a, i.e., a[0],
So what about *(a+i)?
It should represent a reference to the value at index i in array a, i.e., a[i],
So this phenomenon shouldn’t be surprising:
Because *(a+i) == *(i+a)
Therefore a[i] == i[a]

Did you feel an aha! Insight moment?

Extended reading: “C Traps and Pitfalls” p33~p38

Original Chinese version: http://www.cppblog.com/xguru/archive/2009/12/24/103864.html