Hi, I’ve a doubt here. In the Test bench on line 15, it says “+ 8’h01;” but alu_sel is just a 4 bit, so it possible to give 8 bits as input? By the way this is video is excellent and each line explained so well. Thanks for your work!
Have run the code with both " 8'h01 an 4'h01''". It doesn't throw any error and doesn't any change in the output. However in the test bench, whilst intialising the alu_sel to zero (ALU_sel = 4'h0;) it requires a delay it seems(ALU_sel = 4'h0; #10;) if not it just skipping "0000" and goes straight to "0001" and so on.
Sir thank you for this.A very serious and excellent work .!! I hope to see more like this.Thanks again.
best explaination ever ,you deserve more subscribers and views and likes ,
thanks a lot
Hi, I’ve a doubt here. In the Test bench on line 15, it says “+ 8’h01;” but alu_sel is just a 4 bit, so it possible to give 8 bits as input? By the way this is video is excellent and each line explained so well. Thanks for your work!
Have run the code with both " 8'h01 an 4'h01''". It doesn't throw any error and doesn't any change in the output. However in the test bench, whilst intialising the alu_sel to zero (ALU_sel = 4'h0;) it requires a delay it seems(ALU_sel = 4'h0; #10;) if not it just skipping "0000" and goes straight to "0001" and so on.
how this code can be converted in code of 32 bit ALU..??
Give the circuit digram for the same
superb explanation.. could you please send your email id. I have some clarifications regarding verilog. thanks
Where is 5 +4 =9?in output