What's the output of this algorithm?
You can unroll the procedure:

x = 4;
y = 2;
u = x;
v = y;
u = 2*u;
x = u+v;
u = u-1;
print x;


This is semantically equivalent to your program.

Once you copy the value of x to u, the link between the two disappears. This is in contrast to references in C++ or mutable objects in python, which are really pointers; if u were a reference (i.e., if parameters were passed by reference), then the assignment u = x would have identified the two variables as pointing at the same location in memory. As it is, u is just a value (since parameters are passed by value), and so this linkage is absent.