C/C++ Math Order of Operation

Question

So I know that C++ has an Operator Precedence and that

int x = ++i + i++;

is undefined because pre++ and post++ are at the same level and thus there is no way to tell which one will get calculated first. But what I was wondering is if

int i = 1/2/3;

is undefined. The reason I ask is because there are multiple ways to look at that (1/2)/3 OR 1/(2/3). My guess is that it is a undefined behavior but I would like to confirm it.

Answer 1

If you look at the C++ operator precedence and associativity, you'll see that the division operator is Left-to-right associative, which means this will be evaluated as (1/2)/3, since:

Operators that are in the same cell (there may be several rows of operators listed in a cell) are evaluated with the same precedence, in the given direction. For example, the expression a=b=c is parsed as a=(b=c), and not as (a=b)=c because of right-to-left associativity.

Answer 2

In your example the compiler is free to evaluate "1" "2" and "3" in any order it likes, and then apply the divisions left to right.

It's the same for the i++ + i++ example. It can evaluate the i++'s in any order and that's where the problem lies.

It's not that the function's precedence isn't defined, it's that the order of evaluation of its arguments is.

Answer 3

The first code snippet is undefined behaviour because variable i is being modified multiple times inbetween sequence points.

The second code snippet is defined behaviour and is equivalent to:

int i = (1 / 2) / 3;

as operator / has left-to-right associativity.