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:
- From what I learned, signals values are updated only at the end of the process. Is this correct?
- 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?
- 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?
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