blocking assignment in always_comb

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stanford

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1. Are these two code functionally equivalent?

2. Will they synthesize into the same hardware?

Code: 
always_comb
  out = 0;
  out = (sel[0]) ? in[0] : out;
  out = (sel[1]) ? in[1] : out;

Code: 
assign out = sel[1] ? in[1] : sel[0] ? in[0] : 0;
 

FvM said:
What do you expect?


Most likely, if 1 is true.

Code 1 involves a syntax error.
Click to expand...

Are you referring to missing begin/end? I was wondering if they are actually functionally equivalent. I wasn't sure.
 

Assuming the missing begin/end was a typo, they are functionally equivalent and should synthesize to the same hardware. The only difference is in simulation debugging, you'll be able to see the intermediate assignments if you step through the code.
 

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