So 10;3r
will first push 10
onto the stack [10]
, then ;
is a separator between the numbers so that they won't be interpreted as one big number, then 3
push 3
onto the stack [10,3]
. Then when r
otate is called, it takes the top value of the stack (namely the 3
) and uses it as its arguments. This is equivalent to rotate(3);
in Processing code. After the function has been executed, the stack will be [10]
since the 3
has been popped after calling rotate