Does this C++ code produce an undefined behavior? [duplicate]

Question

This question already has answers here:

With arrays, why is it the case that a[5] == 5[a]? (19 answers)

This doesn't look like a function. What is this? (5 answers)

Closed 8 years ago.

A have found the following snippet:

int a[100];
...
int value = 42[a];

Which appears to do exactly what a[42] does.

Is it a bogus with undefined behavior or a perfectly legal C++ code?

Answer 1

It's perfectly legal. With pointer arithmetic, a[42] is equivalent to *(a + 42), which (addition being commutative) is equivalent to *(42+ a), which (by definition of []) is equivalent to 42[a].

So it's obscure, but well-defined.

Answer 2

The array operator is commutative.

a[n] == *(a + n) == *(n + a) == n[a]

And it's perfectly legal.

Answer 3

a[i] is defined as *(a+i).

So 42[a]=a[42]; and it is perfectly safe

Answer 4

42[a] is exactly equivalent to a[42], and entirely legal. It works because a pointer address is just an integer underneath, so you can do the arithmetic either way round (an array variable is really just a pointer in a thin disguise).

It's not usually a good idea for readability though, unless you're deliberately trying to obfuscate the code.