I have translated the following code using ternary. However, I knew there was something wrong with it. Can someone please point me into the right direction?
ForwardA = 0;
ForwardB = 0;
//EX Hazard
if (EXMEMRegWrite == 1) begin
if (EXMEMrd != 0)
if (EXMEMrd == IDEXrs)
ForwardA = 2'b10;
if (EXMEMrd == IDEXrt && IDEXTest == 0)
ForwardB = 2'b10;
end
//MEM Hazard
if (MEMWBRegWrite == 1) begin
if (MEMWBrd != 0) begin
if (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrs)))
if (MEMWBrd == IDEXrs)
ForwardA = 2'b01;
if (IDEXTest == 0) begin
if (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrt)))
if (MEMWBrd == IDEXrt)
ForwardB = 2'b01;
end
end
end
ForwardA = (MEMWBRegWrite && MEMWBrd != 0 && (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrs))) && (MEMWBrd == IDEXrs)) ?
2'b01 : ((EXMEMRegWrite && EXMEMrd != 0 && EXMEMrd == IDEXrs) ? 2'b10 : 0);
ForwardB = (IDEXTest == 0 && MEMWBRegWrite && MEMWBrd != 0 && (!(EXMEMRegWrite == 1 && EXMEMrd != 0 && (EXMEMrd == IDEXrt))) && (MEMWBrd == IDEXrs)) ?
2'b01 : ((EXMEMRegWrite && EXMEMrd != 0 && EXMEMrd == IDEXrt && IDEXTest == 0) ? 2'b10 : 0);
Surprisingly enough, I’m going to risk downvotes and tell you that the right direction is to leave your code in its relatively readable state.
I suspect the only thing you could do that would be worse would be to do it as a regular expression or convert it to inline assembly 🙂
The fact that it’s not converting easily should tell you something about the wisdom in what you’re attempting.
Based on your comment elsewhere:
Well, if you must do it, against my advice (and I’m not alone here in offering this advice), here’s the method you should use (I have no idea what an “always block” even is so I’m not qualified to argue the point with you).
Since your current code is setting
ForwardAandForwardBto values then only changing them under certain conditions, you can transform that into a ternary by reversing the order. That’s because, in yourifversion, later code takes precedence but earlier code takes precedence in the ternary.Find out under what circumstances
ForwardAandForwardBare set in reverse order and reconstruct those conditions.Here’s your original code, compressed a bit. I’ve also changed your
2'b10things into2'b10'so we still get nice formatting in the SO rendering engine – don’t forget to change them back.You can see B is set in three places. It’s set to
2'b01in the bottomif,2'b10in the top one and0at the start. Converting the conditions:Similarly for A:
Now the theory behind that is good but I wouldn’t be the least bit surprised if I’d made an error in the transcription, or if Verilog just threw its hands up in disgust, picked up its ball, and trotted off home 🙂
Can I at least suggest, if you must follow this path, you both:
Seriously, you’ll thank me in six months time when you’re looking over this again, trying to figure out what on Earth you were thinking 🙂