Consider the following code fragment. Notice the use of non-blocking assignment.
always@(posedge clock)
begin
B

Question

Consider the following code fragment.  Notice the use of non-blocking assignment. 
 always@(posedge clock) 
 begin 
   B <= A^C; 
   C <= B; 
 end 
 Which BLOCKING code version would lead to identical logic (after any potential logic gate sharing is taken into account)?

Accepted Answer

always@(posedge clock) 
 begin 
   C = B; 
    B = A^CC; 
 end 
 assign CC=C;

Answer

always@(posedge clock) 
 begin 
   B = A^CC; 
 C = B; 
 end 
 assign CC = C;

Answer

always@(posedge clock) 
 begin 
   C = B;    
   B = A^C; 
 end

Answer

None of the above

Answer

always@(posedge clock) 
 begin 
   B = A^C; 
 C = B; 
 end

Crowdly

Consider the following code fragment. Notice the use of non-blocking assignment...

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