What does the Verilog localparam
X
code below mean? From my understanding it is as follows:
if(y==4) X = +64;
else if(Y==3) X = +32;
else if(Y==2) X = +16;
else if(Y==1) X = +8;
else if(Y==0) X = +4;
else X = +2;
Is this correct? If so, another thing that doesn't make sense to me is the '+' after the '=' sign, e.g., (X = +2)? Is this something unique to localparam
variables?
localparam X = (Y==4) ? +64 :
(Y==3) ? +32 :
(Y==2) ? +16 :
(Y==1) ? +8 :
(Y==0) ? +4 :
+2 ;
2 Answers 2
Yes, the localparam
statement is logically equivalent to the if/else
pseudocode you showed. The statement uses the conditional operator which has precedence like your if/else
.
The +
signs are unnecessary in the code. The code behaves the same as if they were not there. Perhaps someone copy-and-pasted more complicated code (with expressions like A+2
) and forgot to simplify it. A simpler way to write it is:
localparam X = (Y==4) ? 64 :
(Y==3) ? 32 :
(Y==2) ? 16 :
(Y==1) ? 8 :
(Y==0) ? 4 :
2 ;
The conditional operator ?:
is an expression that behaves similar to an if/else
conditional branching procedural statement as you observed. It can be used in places where is is not practical to use procedural statements. A parameter/localparam
must be assigned using an expression. You could wrap the if/else
in a constant function, but the conditional operator is a convenience. If you were making assignments to a variable, a procedural case
statement would be much clearer.
The unary +
operator has no purpose other than symmetry with the unary -
operator. In can make the intent of the code more clear if you have a mix of positive and negative numbers.
+
can be omitted. It is not unique tolocalparam
, it can be anywhere. \$\endgroup\$+
s were added for visual consistency. \$\endgroup\$