Difficulty: Hard · Puzzle 6 of 6 · Topic: randomize() with Puzzles
Predict outcomes of both calls: final len range for Call A and whether post_count increments on Call B.
class life_req;
rand bit [3:0] len;
bit force_short;
int post_count;
constraint c_base { len inside {[1:8]}; }
function void pre_randomize();
if (force_short) len = 2;
endfunction
function void post_randomize();
post_count++;
endfunction
endclass
life_req r = new();
r.force_short = 1;
assert(r.randomize() with { len inside {[5:6]}; }); // Call A
void'(r.randomize() with { len == 9; }); // Call Bpre_randomize runs before solving; post_randomize runs only on success.
1) Call A: pre_randomize assigns len=2, but solver then enforces c_base and inline len in 5..6.
2) So Call A succeeds with len either 5 or 6; tuple-count gives 50/50.
3) post_randomize runs after successful Call A, so post_count increments by 1.
4) Call B asks len==9, conflicting with c_base len 1..8 -> fail.
5) Because Call B fails, post_randomize is not invoked for that call.Answer: Call A succeeds with len in {5,6} (about 50/50), then Call B fails, and post_count increments only once total.
Candidates often think pre_randomize assignment pins the rand value and that post_randomize runs even on failed randomize.
How would you intentionally lock len to 2 for Call A without causing solve failure?