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

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