Verilog Minimum Bit Width

Question

I'm looking for a clean way to declare Verilog/SystemVerilog types with a parameterised bit width. This is what I've got so far and was wondering if there is a better way to do it. I've looked through the system functions in the LRM 1800-2009 and -2017. The closest I could find is $bits, but I would like something like $minbits. Have I overlooked something?

In VHDL, it's done by simply specifying the range:

    signal counter: integer range 0 to MAX_COUNT;

...and the compiler will calculate the minimum bit width to hold that range.

For the parameter values of 20 ns and 125 ms, the counter should be 23 bits with MAX_COUNT being 6,250,000.

module Debounce
#(
    parameter CLOCK_PERIOD_ns    = 20,  // nanoseconds.
    parameter DEBOUNCE_PERIOD_ms = 125  // milliseconds.
)
. . .
    function int MinBitWidth([1023:0] value);
        begin
            for (MinBitWidth = 0; value > 0; MinBitWidth = MinBitWidth + 1)
            begin
                value = value >> 1;
            end
        end
    endfunction

    localparam MAX_COUNT_32BITS = DEBOUNCE_PERIOD_ms * 1_000_000 / CLOCK_PERIOD_ns;  // Default type of 32-bits.

    localparam COUNTER_BITS = MinBitWidth(MAX_COUNT_32BITS);  // Calculate actual bit width needed.
    typedef logic [COUNTER_BITS - 1 : 0] TCounter;
    localparam TCounter MAX_COUNT = MAX_COUNT_32BITS;  // Assign to a type of the actual bit width (truncation warning from Quartus).
    localparam TCounter ONE = 1;
    TCounter counter;
. . .
    always @(posedge clock)
    begin
. . .
        if (counter == MAX_COUNT_32BITS - 1)  // Synthesises a 32-bit comparer no matter how many bits are needed with unused bits tied to ground.
. . .
        if (counter == MAX_COUNT - ONE)       // Synthesises a 23-bit comparer as expected.
. . .
            counter <= counter + 1;           // Synthesises a 23-bit counter as expected.
. . .
            counter <= counter + ONE;         // Synthesises a 23-bit counter as expected.

Incorrect Algorithm

I considered $clog2 which is the correct way to obtain an address bus width from a RAM depth parameter. However, this is not the same as the minimum bit width of a value. Let me explain...

Consider a value of 4 which is 100 base-2 (3 bits wide).

The $clog2 algorithm calculates a value of 2, which is incorrect. It should be 3. The reason for this miscalculation is because $clog2 subtracts 1 from the value before it starts to compute the number of bits, i.e. 4 becomes 3, then it calculates the minimum bit width of the value 3, giving 2 bits. While this is mathematically correct for the ceiling of log base-2, it is not the bit width of the original value.

Here is the clogb2 algorithm from the LRM:

function integer clogb2;
    input [31:0] value;
    begin
        value = value - 1;  // GOTCHA!
        for (clogb2 = 0; value > 0; clogb2 = clogb2 + 1) begin
            value = value >> 1;
        end
    end
endfunction

Correct Algorithm

The correct algorithm is to calculate the minimum bit width of the original value, which is the algorithm given by @jonathan-mayer in his first answer before he edited it.

Here is the correct algorithm as a function:

function integer MinBitWidth;
    input [1023:0] value;
    begin
        for (MinBitWidth = 0; value > 0; MinBitWidth = MinBitWidth + 1)
        begin
            value = value >> 1;
        end
    end
endfunction

score 1 · Answer 1 · edited Jun 24 '20 at 08:39

1

Just do +1 to get correct values for powers of 2.

$clog2(MAX_COUNT_32BITS + 1);

edited Jun 24 '20 at 08:39

tim

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answered Jun 21 '20 at 08:30

Moberg

5,253
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That's a great suggestion to fix the algorithm, but it kind of obfuscates the meaning of it from a semantic point of view. I could replace the body of the `MinBitWidth` function with `$clog2(value+1);` to keep the semantic meaning, but I'm also looking for a way to eliminate the rest of the awkward coding. – tim Jun 21 '20 at 15:58
Set the count to +1. This is the natural range, because it includes zero. If the maximum value is 4, then you need space to support 5 values. [0 .. 4]. There is no 'un-naturalness' to this answer. It is a fact of zero based numbers. The define is semantically wrong. It is MAX_COUNT_VALUE as '4' and MAX_COUNT_RANGE as '5'. – artless noise Jul 11 '23 at 23:38

score 0 · Answer 2 · answered Jun 21 '20 at 00:39

0

$clog2 from IEEE Std 1800-2017, section 20.8.1 Integer math functions:

The system function $clog2 shall return the ceiling of the log base 2 of the argument (the log rounded up to an integer value).

module tb;

    parameter CLOCK_PERIOD_ns    =  20;  // nanoseconds.
    parameter DEBOUNCE_PERIOD_ms = 125;  // milliseconds.
    localparam MAX_COUNT_32BITS = DEBOUNCE_PERIOD_ms * 1_000_000 / CLOCK_PERIOD_ns;  // Default type of 32-bits.
    localparam COUNTER_BITS = $clog2(MAX_COUNT_32BITS);  // Calculate actual bit width needed.

initial begin
    $display("MAX_COUNT_32BITS = %0d, COUNTER_BITS = %0d", MAX_COUNT_32BITS, COUNTER_BITS);
end

endmodule

Outputs:

MAX_COUNT_32BITS = 6250000, COUNTER_BITS = 23

answered Jun 21 '20 at 00:39

toolic

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I considered that after reading your answer about ADDRESS BUS WITDH (https://stackoverflow.com/questions/5269634/address-width-from-ram-depth/)[here], but there is a very subtle (even crafty) reason why it won't work. I'll update my post to explain the difference between min bit width and address bus width. – tim Jun 21 '20 at 00:44
I've updated it now to show how `$clog2` can't calculate the minimum bit width of powers of 2. – tim Jun 21 '20 at 01:19
Yes, but it's awkward coding. I would like to know if Verilog/SystemVerilog has this algorithm built-in somewhere so that I can use much simpler code like VHDL. – tim Jun 21 '20 at 01:31
2

@tim Since you know that $clog2 subtracts one and you don't want that is there a reason that you don't use $clog2(value + 1)? – Brad Budlong Jun 21 '20 at 04:29
That's a great suggestion to fix the algorithm within the function `MinBitWidth`, but I'm looking for a way to eliminate the rest of the awkward coding as well. – tim Jun 21 '20 at 15:59

Verilog Minimum Bit Width

Incorrect Algorithm

Correct Algorithm

2 Answers2