Question

2. Evaluate these two pieces of code if Q starts with a value of Q [4:0] = 11001. Show work so that it is obvious that you understand. Q[3] = Q [4]; Q [2] = Q [3]; Q[4] = 1'60; Q [0]=Q [1]; Q [1]=Q [2]; Q[3] <= Q [4]; Q [2] <= Q [3]; Q [4] <= 1'b0; Q [0] <= Q [1]; Q [1] <= Q [2};

Answer 1

a values. Q043=1; 913] =15, QC23-0QCIJ=0 case o Here case ☺ TS non-blocking assignment that means non- blocking assignments are executed concurrently with an the sequentially blocks that .TS a non- blocking assignment execotes without waiting for the non-blocking assign -ments occuring at the same time to complete. register <= expression; inital Cise Qro] =1; Q.C3]<: ocul; C2] <= QC3]; → hot update It takes Intial value be caus QC4] c= ibo, does't wait for the above result ♡co] C: 0013; Q CI] = Q(2]; olpi - 0 [3] = 1 ; 0623:1? C] = 1; QC 03-02 Q[i]=0;

Qu Q3 o=1 QC3] <.004).; Q6:23.<= Q(3] Q(4)=1; [3] =1 ? Q[2] =]QC 1] = 1; Given that Q14:0] - 11001 Q4=1, Q3= 17,020, 0,30; casei caseir Q13] - QC4]; QP2] = Q [3]; Q[4] Q[o] = Q [1] QCI] = Q(23) Case ū (st waits for executed Q(4)=1. cuseſ Finally op's? - ibo Qru] <a Q[o] ca ibo; QC13; Q[2]; Q Cl]<a the above Q (3) = 1 expression execution) (3): 1; « C2) = Q(3) = 1; (update) > It is a procedural manner (serial manner) QCO=0; Q (1):1 W CO] =0 case in has above change and . In both case is the same TS some not outputs only but case → case ů :PS Blocking assignment that means in the order Blocking assignments executed are the sequen tiw that block specified TS a blocking assignment waits for Perious blocking ässignment of the to Same complete before executing time . expression register