Operator precedence for And/&& in Ruby [duplicate]

Answer 1

The reason is simple: precedence. As you say, the order is:

1. &&
2. =
3. and

Since && has precedence over =, the statement is evaluated like this:

if a = (f(2) && (b = f(4))) then 

Which results in:

if a = (2 && 4) then

When x and y are integers, x && y returns y. Thus 2 && 4 results in a = 4.

For comparison's sake, the first one is evaluated like this:

if (a = f(2)) and  (b = f(4)) then