Negative array index
access-violation, arrays, c++, pointers
Solution
`X[Y]` is identical to `*(X + Y)` as long as one of `X` and `Y` is of pointer type and the other has integral type. So `b[-1]` is the same as `*(b - 1)`, which is an expression that may or may not be evaluated in a well-formed program – it all depends on the initial value of `b`! For example, the following is perfectly fine:
int q[24];
int * b = q + 13;
b[-1] = 9;
assert(q[12] == 9);
In general, it is your responsibility as a programmer to guarantee that pointers have permissible values when you perform operations with them. If you get it wrong, your program has undefined behaviour. For example:
int * c = q; // q as above
c[-1] = 0; // undefined behaviour!
Finally, just to reinforce the original statement, the following is fine, too:
std::cout << 2["Good morning"] << 4["Stack"] << 8["Overflow\n"];
Problem
I have a pointer which is defined as follows: ``` A ***b; ``` What does accessing it as follows do: ``` A** c = b[-1] ``` Is it an access violation because we are using a negative index to an array? Or is it a legal operation similar to `*--b`? EDIT Note that negative array indexing has different support in C and C++. Hence, this is not a dupe.