CODEWITHSUNDEEP
verilog
Dragonsheep
Dragonsheep

Reputation: 239

Evaluation order for always blocks triggered within always blocks in Verilog?

I understand that, for 2 always blocks with the same trigger, their order of evaluation is completely unpredictable.

However, suppose I have:

always @(a) begin : blockX 
c = 0;
d = a + 2;
if(c != 1) e = 2;
end

always @(a) begin : blockY 
e = 3;
end

always @(d) begin : blockZ 
c = 1;
e = 1;
end

Suppose block X evaluates first. Does changing d in blockX immediately jump to blockZ? If not, when is blockZ evaluated with respect to blockY?

My programmer's instinct thinks of the sequence of events as a stack, where evaluating blockX is like a function call to blockZ and I immediately jump there in the code, then finish evaluating blockX.

However, because we call the active events queue, well, a queue, this suggests blockZ is enqueued at the back of the active events queue, and I'm 100% guaranteed it will be evaluated last (unless there are other triggered always blocks).

There's also the intermediate possibility, where it's neither first nor last but is also evaluated in a random and unpredictable order.

So in this example, are 1, 2, or 3 all possible final values for e, depending on how the compiler is feeling at run time?

Additionally, while I understand, of course, this represents awful style, where might I find the specification for this kind of behvaior?

Upvotes: 1

Views: 585

Answers (1)

dave_59
dave_59

Reputation: 42623

Always blocks are not function calls. See a recent answer I just gave for a similar question. These blocks are concurrent processes. The LRM only guarentees the ordering of statements within a begin/end block. There is no defined ordering between concurrently executing begin/end blocks (See Section 4.7 Nondeterminism in the 1800-2012 LRM) So a simulator is free to interleave the statements in any way as long as it honors the order within a single block.

So you are correct that e could have the final values 1, 2 or 3 depending on how a simulator decides to implement and optimize your code.

Upvotes: 1

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