How does signal assignment work in a process?

问题:

I learned that a signal is not changed immediately when encountering an expression, but when the process ends. In this example here:

...
signal x,y,z : bit;
...
process (y)
begin
  x<=y;
  z<=not x;
end process;

The example says this:

If the signal y changes then an event will be scheduled on x to make it the same as y. Also, an event is scheduled on z to make it the opposite of x. The question is, will the value of z be the opposite of y? Of course, the answer is no, because when the second statement is executed, the event on x has not been processed yet, and the event scheduled on z will be the opposite of the value of x before the process begins.

Well, I need to understand some things:

1. From what I learned, signals values are updated only at the end of the process. Is this correct?
2. The signal x is updated as the first statement. This does not still change the value of x, this change is put in a queue to be executed after the process ends. So everything after this statement x <= y will not see the change and will see x having its old value. Is this correct?
3. The second statement is an attempt to change the value of signal z. The same here, z will not change its value but it depends on the value of another process. The change on z will be put in queue to be executed at the end of the process. Is this correct?

What does happen at the end of the process?

Possibility number 1) The value in x is changed so x has its new value. The second signal z is updated, the first signal x was updated and, given that z depends on x, its value is changed basing on the NEW UPDATED value of x. And the example should work fine.

Possibility number 2) The value in x is changed so x has its new value. The second signal z is updated. Given that z was assigned an old value of x, that's the value that z will hold, the old value of x which was updated, but this update is not considered.

Could you please tell me which one is the correct way?

回答1:

Variables get updated as you assign them. Signals get update in the next delta cycle (at the earliest).

a := '1'; -- variable
assert a = 1;
b <= '1'; -- signal
computationUsing(b); --reads old value of b
-- new value will be visible after this process ends or some time passes

Jan Decaluwe explains this stuff in more detail here: http://www.sigasi.com/content/vhdls-crown-jewel

回答2:

The way it works:

Y changes and the process begins.

X will be assigned to what Y's value currently is, but not until the end of the process

Z will be assigned to not X's old value but not until the end of the process

The process ends so now X and Z will be updated

回答3:

I disagree with Ashraf's post. I have made vhdl code myself where variables are wires, and signals are latches. Examples:

signal x,y,clk; process(clk) begin x <= y end process

This creates a synchronous latch, a flip flop.

Any variable which does not assign its value to a signal, but only to other variables, is a perfectly acceptable "wire".

My understanding of the whole subject is this:

A signal assignment inside a process will disregard other signal assignments made in the same process "instantiation". Also, for the same signal, only the last assignment will be taken into account.

About "Ok END OF THE PROCESS: What does it happen?????":

I think that a signal assignment will take place at the fastest possible time the hardware utilization of the process allows. EXCEPTION: Changes within a if(rising_edge(clk)) will take place at the start of the next clock cycle.