0
\$\begingroup\$

I know that if I use

s1 |=> s2

the consequent sequence expression s2 will be evaluated on the next clock tick. However, if there is no clock(combinational circuit) how will the assertion be evaluated?Will the simulator wait for 1 time period and check s2 or is it something else?

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

There needs to be a clock declared somewhere. It can be declared in the sequence itself, the property calling the sequence, the assertion calling the property/sequence, or a default clocking block in the module/interface where the assertion/property/sequence resides.

You don't usually put assertions on conbinational logic. With RTL combinational logic you usually should not consider delay. Assertions are intended for flop to flop checking.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ I wouldn't make that strong of a recommendation against using assertions in combinatorial logic. You might want to assert that a signal or set of signals maintain some legal set of values or relationships over all time without a clock (like one-hot). But you are correct to say that any assertion that has a cycle delay needs to have a defined clock. \$\endgroup\$
    – dave_59
    Mar 17, 2015 at 23:19
  • \$\begingroup\$ My recommendation to limit immediate assertions in RTL comb-blocks is base on: (1) they disappears in gate simulations and (2) may give false errors for zero-time glitches w/ noisy inputs and will degrades simulation performance with assertion coverage collection enabled. Usually, my one-hots go to a unique case statement; giving me a one-hot check that doesn't report zero-time glitches as errors, plus gives the intended full_case/parallel_case directive for synthesis. There is a place for assertions in comb-logic, my recommendation is to use them sparingly in RTL. \$\endgroup\$
    – Greg
    Mar 18, 2015 at 18:07
  • \$\begingroup\$ Not really logic assertions, but you might want to put un-clocked assertions on parameter values to make sure they are in an expected range. \$\endgroup\$
    – dave_59
    Mar 19, 2015 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.