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Suppose one has an entity which has two architectures defined. Those two architectures work with the same entity (obviously) and subsequently the two set the output pins to different values. My question is, how does the program (simulator) determine what the output should be (i.e. which architecture to choose)?

Here is an example:

library ieee;
use ieee.std_logic_1164.all;

entity Exercise_4 is 
generic (n : integer := 4);
port(
a, b : std_logic_vector (n-1 downto 0);
clk, rst : std_logic;
q, qn : buffer std_logic_vector (n-1 downto 0));
end;

architecture one of Exercise_4  is
begin
process (clk, rst)
    begin
    if rst = '0' then 
        q <= (others=>'0');
    elsif (clk' event and clk = '0') then
        q <= a ;
    end if;
end process;

process (clk, rst)
begin
    if rst = '0' then 
        qn <= (others=>'1');
    elsif (clk' event and clk = '0') then
        for i in a'range loop
            qn(i) <= not q(i) ;
        end loop;
    end if;
end process;
end;

architecture two of Exercise_4  is
begin
process (clk,rst)
    begin
    if rst = '0' then 
        q <= (others=>'0'); 
        qn <= (others=>'0');
    elsif (clk' event and clk = '0') then
        q <= a;
        qn <= b ;
    end if;
end process;
end;

I did a simulation and saw that q gets the value of a assigned and qn gets the value of b assigned. It seems that the second architecture has been chosen by the compiler I don't understand why the program decided to do so.

Thank you.

Jim McAdams
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nettek
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    It depends on your compiler. Some are just taking the first arch, some are using the dedicated [configuration](https://www.doulos.com/knowhow/vhdl_designers_guide/configurations_part_1/) construct. – Eugene Sh. Apr 13 '15 at 20:31
  • Weird. As you can see I did not even use a configuration construct. And the compiler "mixed" the two architectures - it took `q` from the first and `qn` from the second. Is that possible? – nettek Apr 13 '15 at 21:48
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    Eugene's comment is not accurate for a compliant VHDL implementation. In the absence of a binding indication a default binding indication will use "the most recently analyzed architecture body associated with the entity declaration" - the second appearing in the design file or last analyzed architecture. See IEEE Std 1076-2008 7.3.3 Default binding indication, paragraph 4. (You may need to read most of 7.3). And no taking `q` from one architecture and `qn` from the other isn't possible, which also says you haven't demonstrated your problem adequately. –  Apr 14 '15 at 01:25
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    In both architecture, `q` is assigned to `a`. I don't see why `q` would be any different in the two architectures. `qn` will though. – Jonathan Drolet Apr 14 '15 at 01:56
  • Yes you're both right @JonathanDrolet @DavidKoontz , `q` gets `a` from the second architecture as well. I was confused as the same line is in both architectures, and I failed to notice (it was late at night as well). Anyway, Thank you everyone. – nettek Apr 14 '15 at 15:33

2 Answers2

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If you don't specify yourself which architecture to choose² then the compiler will take "the most recently analyzed architecture body associated with the entity declaration" (assuming the compiler is compliant to the IEEE standard) [1].

² You can select the architecture you prefer e.g. in the component declaration section (where you map the signals) on a higher design level:

entity topentity is 
end;

architecture toparch of topentity is

  -- component instantiation
  component Exercise_4 is
  generic (n : integer := 4);
  port(
    a, b : std_logic_vector (n-1 downto 0);
    clk, rst : std_logic;
    q, qn : buffer std_logic_vector (n-1 downto 0));
  end component Exercise_4;

begin

  -- component mapping
  E4: entity work.Exercise_4(one)
  generic map ( .. )
  port( .. );

end architecture toparch;

[1] IEEE Std 1076-2008 7.3.3 Default binding indication, paragraph 4.

Disclaimer: Answer was constructed with the help of the comments above. No copyright infringement intended. ;P

Jim McAdams
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0

In absence of configuration binding, by default the compiler consider the last architecture body in the code file.

Dr. Jitendra Chourasia
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