Verilog question mark (?) operator

Question

I'm trying to translate a Verilog program into VHDL and have stumbled across a statement where a question mark (?) operator is used in the Verilog program.

The following is the Verilog code;

1  module music(clk, speaker);
2  input clk;
3  output speaker;
4  parameter clkdivider = 25000000/440/2;

5  reg [23:0] tone;
6  always @(posedge clk) tone <= tone+1;

7  reg [14:0] counter;
8  always @(posedge clk) if(counter==0) counter <= (tone[23] ? clkdivider-1 : clkdivider/2-1); else counter <= counter-1;

9  reg speaker;
10  always @(posedge clk) if(counter==0) speaker <= ~speaker;
11  endmodule

I don't understand the 8th line, could anyone please shed some light on this? I've read on the asic-world website that the question mark is the Verilog alternate for the Z character. But I don't understand why it's being used in this context.

Kind regards

Answer 1

That's a ternary operator. It's shorthand for an if statement

Format:

condition ? if true : if false

Example:

tone[23] ? clkdivider-1 : clkdivider/2-1

Translates to something like (not correct syntax but I think you'll get it):

if tone[23] is 1, counter = clkdivider-1
else counter = clkdivider/2-1

Here are two examples of a 2 to 1 MUX using if statement and ternary operator.

On the asic-world website, it is covered under Conditional Operators

Answer 2

Another way of writing, e.g. the following Verilog:

q <= tone[23] ? clkdivider-1 : clkdivider/2-1;

in VHDL would be:

q <= clkdivider-1 when tone[23] else clkdivider/2-1;